Chapter 7.6 - Formula for Number of Combinations

In the previous section, we saw a relation between number of permutations and number of combinations of n objects taken r at a time.

• We analyzed two examples. In both of those examples, we obtained the same result.
• In both of those examples, the number of objects inside the combinations were ‘2’.
• Let us see another example in which the number of objects inside the combinations is 3. It can be written in 4 steps:
1. Consider five objects A, B, C, D and E
• We want to make combinations of these five objects, taking 3 at a time.
• Some of the possible combinations are: ABC, BCD, ADE, . . .
2. We want to know how many such combinations are possible.
• In other words, we want to calculate ${}^4 C_3$.
3. Consider the combination ABC. In this combination, the order is not important.
• But the three objects A, B and C can be arranged among themselves in 3! ways.
◼ So we can write:
Number of combinations × 3! = Number of permutations
4. Now we get the same idea as before:
   ♦ Number of combinations of n objects taking r at a time $\left({}^n C_r \right)$
   ♦ Multiplied by
   ♦ Number of permutations inside each combination $(r!)$
   ♦ Will give
   ♦ The number of permutations of n objects taking r at a time $\left({}^n P_r \right)$
• So here also, we can write: ${}^n C_r~ × ~ r!~=~{}^n P_r$

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• We get the same relation in all three examples.
• We can confirm that the relation between ${}^n C_r$ and ${}^n P_r$ is: ${}^n C_r~ × ~ r!~=~{}^n P_r$
• Now, we already have the formula: ${}^n P_r~=~\frac{n!}{(n-r)!}$
• So the relation becomes: ${}^n C_r~ × ~ r!~=~\frac{n!}{(n-r)!}$
• This can be rearranged as: ${}^n C_r~=~\frac{n!}{r!~ × ~(n-r)!}$
• Thus we get the formula for ${}^n C_r$

Let us see some important results related to the above formula

Result 1:
This can be written in 3 steps:
1. ${}^n C_r$ is the number of combinations when n objects are taken r at a time.
• Now, instead of r, what if we take all the n objects?
2. Using the formula, we can write:
${}^n C_r~=~{}^n C_n~=~\frac{n!}{n!~ × ~(n-n)!}~=~\frac{n!}{n!~ × ~0!}~=~\frac{n!}{n!~ × ~1}~=~1$
3. That means, if we take all objects at a time, there is only one combination possible.

Result 2:
This can be written in 3 steps:
1. ${}^n C_r$ is the number of combinations when n objects are taken r at a time.
• Now, instead of r, what if we take zero objects?
2. Using the formula, we can write:
${}^n C_r~=~{}^n C_0~=~\frac{n!}{0!~ × ~(n-0)!}~=~\frac{n!}{1~ × ~n!}~=~1$
3. Taking zero objects means that, we are leaving behind all the n objects.
All the n objects can have a combination of only 1.

Result 3:
This can be written in 2 steps:
1. Based on the results 1 and 2, we can say that r can be equal to both zero or n.
2. So we can write the formula as:
${}^n C_r~=~\frac{n!}{r!~ × ~(n-r)!}~~0 \le r \le n$

Result 5:
This can be written in 4 steps:
1. We have the formula: ${}^n C_r~=~\frac{n!}{r!~ × ~(n-r)!}$
Here, r objects are taken.
2. What if we take (n-r) objects?
We get: ${}^n C_{n-r}~=~\frac{n!}{(n-r)!~ × ~[n-(n-r)]!}~=~\frac{n!}{(n-r)!~ × ~[n-n+r]!}~=~\frac{n!}{(n-r)!~ × ~r!}$
$\Rightarrow {}^n C_{n-r}~=~\frac{n!}{(n-r)!~ × ~r!}$
3. Now compare the expressions in (1) and (2)
• We see that, the numerators and denominators are the same.
• Thus we get: ${}^n C_r~=~{}^n C_{n-r}$
4. So we can write:
   ♦ We may take r objects at a time.
   ♦ Or we may reject r objects and take the remaining (n-r) objects at a time.
   ♦ In both cases, the number of combinations will be the same.

Result 6:
This result is based on the previous result 5. It can be written in 3 steps:
1. Suppose that, ${}^n C_a~=~{}^n C_b$
2. Then there are two possibilities:
• Possibility (i): a = b
• Possibility (ii) Based on result 5, we can write: b = n-a
    ♦ This gives: n = a+b
3. So we can write:
If ${}^n C_a~=~{}^n C_b$, then a = b or n = a+b

Result 7:
${}^n C_r~+~{}^n C_{r-1}~=~{}^{n+1} C_r$
• Proof can be written in 3 steps:
1. The LHS can be simplified as:
$\begin{array}{ll}
{}&{}^n C_r~+~{}^n C_{r-1}&{}={}&\frac{n!}{r!~ × ~(n-r)!}~+~\frac{n!}{(r-1)!~ × ~[n-(r-1)]!}&{} \\
{}&{}&{}={}& \frac{n!}{r!~ × ~(n-r)!}~+~\frac{n!}{(r-1)!~ × ~(n-r+1)!}&{} \\
{}&{}&{}={}& \frac{n!}{r(r-1)!~ × ~(n-r)!}~+~\frac{n!}{(r-1)!~ × ~(n-r+1)!}&{\color {green}{\because ~r!=r(r-1)!}} \\
{}&{}&{}={}& \frac{n!}{r(r-1)!~ × ~(n-r)!}~+~\frac{n!}{(r-1)!~ × ~(n-r+1)(n-r)!}&{\color {green}{\because ~(n-r+1)!=(n-r+1)(n-r)!}} \\
{}&{}&{}={}& \frac{n!}{(r-1)!~ × ~(n-r)!}~\left[\frac{1}{r}~+~\frac{1}{(n-r+1)}\right]&{} \\
{}&{}&{}={}& \frac{n!}{(r-1)!~ × ~(n-r)!}~\left[\frac{n-r+1+r}{r(n-r+1)}\right]&{} \\
{}&{}&{}={}& \frac{n!}{(r-1)!~ × ~(n-r)!}~\left[\frac{n+1}{r(n-r+1)}\right]&{} \\
{}&{}&{}={}& \frac{[(n+1)n!]}{[r(r-1)!]~[(n-r+1)(n-r)!]}&{} \\
{}&{}&{}={}& \frac{[(n+1)!]}{[r(r-1)!]~[(n-r+1)(n-r)!]}&{\color {green}{\because ~(n+1)n!=(n+1)!}} \\
{}&{}&{}={}& \frac{[(n+1)!]}{[r!]~[(n-r+1)(n-r)!]}&{\color {green}{\because ~r(r-1)!=r!}} \\
{}&{}&{}={}& \frac{[(n+1)!]}{[r!]~[(n-r+1)!]}&{\color {green}{\because ~(n-r+1)(n-r)!=(n-r+1)!}} \\
{}&{}&{}={}& \frac{(n+1)!}{r!~(n+1-r)!}&{\color {green}{\because ~(n-r+1)(n-r)!=(n-r+1)!}} \\
\end{array}$
2. The RHS can be simplified as:
$\begin{array}{ll}
{}&{}^{n+1} C_r&{}={}&\frac{(n+1)!}{r![(n+1)-r]!}&{} \\
{}&{}&{}={}&\frac{(n+1)!}{r!(n+1-r)!}&{} \\
\end{array}$
3. Thus we get: LHS = RHS

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Now we will see some solved examples

Solved example 7.18
If ${}^n C_9~=~{}^n C_8$, then find ${}^n C_17$
Solution:
1. We have result 6:
If ${}^n C_a~=~{}^n C_b$, then a = b or n = a+b
2. In our present case, 9 cannot be equal to 8.
• So the only option is: n = 9+8
• Thus we get: n = 17
3. So ${}^n C_{17}~=~{}^{17} C_{17} ~=~1$ (using result 1)

Solved example 7.19
A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?
Solution:
Part (i):
1. Let the two men be denoted as M₁ and M₂
• Let the three women be denoted as W₁, W₂ and W₃
2. Then one possibility is: M₁, W₁, W₃
• Another possibility is: W₂, M₂, M₁
• There are many possibilities like this.
3. But in this problem, order is not important because all three members will be having equal status. We need not denote them as M₁, M₂ etc.,
• It is a combination of 5 people taking 3 at a time.
4. So we get:
Number of combinations = ${}^5 C_3~=~\frac{5!}{3! (5-3)!}~=~10$

Part (ii):
1. Out of the 10 possible combinations, some of them will have exactly one man and two women. We want to find the number of such combinations.
2. Assigning units:
• In fig.7.8 below, the first box is considered as one unit. It is for men.
• The remaining two boxes together are considered as one unit. It is for women.

Fig.7.8

3. Filling the units:
• The first unit can be filled in ²C₁ = 2 ways
• The second unit can be filled in ³C₂ = 3 ways
4. So by applying the multiplication principle, the two units can be filled in (2 × 3)= 6 ways
5. We can write:
Out of the 10 combinations, 6 will have exactly one man and two women.

Solved example 7.20
What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these
(i) four cards are of the same suit,
(ii) four cards belong to four different suits,
(iii) are face cards,
(iv) two are red cards and two are black cards,
(v) cards are of the same color?
Solution:
• A pack of 52 playing cards will have four suits. Each suit will have 13 cards. Details can be seen here.
• We can write:
4 cards can be chosen in ⁵²C₄ = 270725 ways.
Part (i): Four cards are of the same suit
1. Let us assume that, all 4 cards are diamonds.
Four diamond cards can be chosen from 13 diamond cards in ¹³C₄ ways.
2. Let us assume that, all 4 cards are clubs.
Four clubs cards can be chosen from 13 clubs cards in ¹³C₄ ways.
3. Let us assume that, all 4 cards are spades.
Four spades cards can be chosen from 13 spades cards in ¹³C₄ ways.
4. Let us assume that, all 4 cards are hearts.
Four hearts cards can be chosen from 13 hearts cards in ¹³C₄ ways.
5. So the total number of ways = 4 × ¹³C₄ ways = 2860

Part (ii): Four cards belong to four different suits
1. Imagine that there are four boxes.
2. The first box can be filled in = 52 ways
• If the selected one is a diamond, no diamond should be used for filling the remaining 3 boxes.
    ♦ So there will be only (52-13) = 39 cards available to fill the second box.
• If the selected one is a club, no club should be used for filling the remaining 3 boxes.
    ♦ So there will be only (52-13) = 39 cards available to fill the second box.
• If the selected one is a spade, no spade should be used for filling the remaining 3 boxes.
    ♦ So there will be only (52-13) = 39 cards available to fill the second box.
• If the selected one is a heart, no heart should be used for filling the remaining 3 boxes.
    ♦ So there will be only (52-13) = 39 cards available to fill the second box.
3. The second box can be filled in = 39 ways
• If the selected one is a diamond, no diamond should be used for filling the remaining 2 boxes.
    ♦ So there will be only (39-13) = 26 cards available to fill the third box.
• If the selected one is a club, no club should be used for filling the remaining 2 boxes.
    ♦ So there will be only (39-13) = 26 cards available to fill the third box.
• If the selected one is a spade, no club should be used for filling the remaining 2 boxes.
    ♦ So there will be only (39-13) = 26 cards available to fill the third box.
• If the selected one is a heart, no heart should be used for filling the remaining 2 boxes.
    ♦ So there will be only (39-13) = 26 cards available to fill the third box.
4. Based on this, we can write:
    ♦ Box 1 can be filled in 52 ways
    ♦ Box 2 can be filled in 39 ways
    ♦ Box 3 can be filled in 26 ways
    ♦ Box 4 can be filled in 13 ways
5. So the total number of permutations is: (52 × 39 × 26 × 13) = 685464 ways.
• But order is not important. The four cards can be arranged among themselves in 4! ways.
• Thus the number of combinations = $\frac{685464}{4!}$ = 28561 ways.

Alternate method:
1. Imagine that there are four boxes.
2. Let us fill the first box with a diamond card.
• A diamond card can be chosen from 13 diamond cards in 13 ways.
3. Let us fill the second box with a spade cards.
• A spade card can be chosen from 13 spade cards in 13 ways.
4. Let us fill the third box with a club cards.
• A club card can be chosen from 13 club cards in 13 ways.
5. Let us fill the fourth box with a heart cards.
• A heart card can be chosen from 13 heart cards in ways.
6. So by applying the multiplication principle,
the four boxes can be filled in (13 × 13 × 13 × 13) = 28561 ways

• This example helps us to understand the relation between permutation and combination.

Part (iii): four cards are face cards
1. In any one suit, there are 3 face cards. So in the whole pack, there will be (4 × 3) = 12 cards
2. From these 12 cards, four cards can be selected in ¹²C₄ = 495 ways

Part (iv): two are red cards and two are black cards
1. Imagine that there are four boxes.
    ♦ First two boxes can be considered as one unit.
    ♦ The remaining two boxes can be considered as another unit.
This is shown in fig.7.9 below:

Fig.7.9

 
2. Let us fill the first unit with red cards.
    ♦ There are 26 red cards.
    ♦ So the two boxes of the first unit can be filled in ²⁶C₂  ways
3. Let us fill the second unit with black cards.
    ♦ There are 26 black cards.
    ♦ So the two boxes of the second unit can be filled in ²⁶C₂  ways
4. So by applying the multiplication principle, the two units can be filled in
(²⁶C₂  × ²⁶C₂) = 105625 ways

Part (v): cards are of the same color ?
1. Imagine that there are four boxes.
2. First let us fill all the four boxes with red cards.
    ♦ There are 26 red cards. So 4 cards can be chosen in ²⁶C₄  ways.
3. Next let us fill all the four boxes with black cards.
    ♦ There are 26 black cards. So 4 cards can be chosen in ²⁶C₄  ways.
4. So in total, there will be (2 × ²⁶C₄) = 29900 ways.

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The link below gives some more solved examples:

Exercise 7.4

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We have completed a discussion on combinations. In the next section, we will see some miscellaneous examples.

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Chapter 7.4 - Permutations When All Objects Are Not Different

In the previous section, we derived various formulas using the factorial notation. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 7.11
How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
Solution:
We have formula A: ${}^nP_r=\frac{n!}{(n-r)!}$
   ♦ Here n = 9 and r = 4
• So we get:
${}^nP_r={}^9P_4=\frac{9!}{(9-4)!}=\frac{9!}{5!}=\frac{9 × 8 × 7 × 6 × 5!}{5!}=9 × 8 × 7 × 6=3024$

Solved example 7.12
How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed?
Solution:
1. There is a series of numbers between 100 and 1000
• The series starts at 101 and ends at 999
    ♦ All members of this series are 3-digit numbers.
    ♦ In fact, all 3-digit numbers except 100 will be a member of this series.
2. So we have to answer this question:
How many 3-digit numbers can be made using the six digits: 0, 1, 2, 3, 4 and 5 ?
3. There are 6 digits available. We take 3 at a time.
• So the number of permutations = ${}^nP_r={}^6P_3=\frac{6!}{(6-3)!}=\frac{6!}{3!}=\frac{6 × 5 × 4 × 3!}{3!}=6 × 5 × 4=120$
4. But out of those 120 permutations, some will have '0' in the 100's place.
• Such numbers cannot be allowed because, 3-digit numbers with 0 at the 100's place are actually 2-digit numbers.
5. So our next task is to find the number of permutations, which have 0 at the 100's place. 
• For that, we put 0 in the first box.
• The remaining two boxes can be filled in (5 × 4) = 20 ways.
6. So out of the 120 permutations, 20 will have 0 at the 100's place.
• Thus the number of 3-digit numbers using the given digits = (120-20) = 100

Solved example 7.13
Find the value of n such that,
(i) ${}^nP_5=42({}^nP_3),~n>4$ (ii) $\frac{{}^nP_4}{{}^{n-1}P_4}=\frac{5}{3},~n>4$
Solution:
Part (i):
1. ⁿP₅ is the permutation of n objects taken 5 at a time.
• Recall that, before learning the factorial notation, we had a lengthy equation:
ⁿP_r = n(n-1)(n-2)(n-3) . . . (n-r+1)
• So we can write:
ⁿP₅ = n(n-1)(n-2)(n-3) . . . (n-5+1)
⇒ ⁿP₅ = n(n-1)(n-2)(n-3)(n-4)
2. Similarly, ⁿP₃ = n(n-1)(n-2)
3. So the given equation becomes:
n(n-1)(n-2)(n-3)(n-4) = 42 × n(n-1)(n-2)
$\Rightarrow~\frac{n(n-1)(n-2)(n-3)(n-4)}{n(n-1)(n-2)}=42$
4. Given that n > 4
• So (n-1), (n-2), (n-3) and (n-4) are all greater than 0.
• So the ratio $\frac{n(n-1)(n-2)(n-3)(n-4)}{n(n-1)(n-2)}$ exists.
• We get: $\frac{(n-3)(n-4)}{1}=42$
$\Rightarrow~n^2-4n-3n+12=42$
$\Rightarrow~n^2-7n=30$
5. This is a quadratic equation. We can solve it using square completion method.
$n^2-7n+\left( \frac{7}{2} \right)^2=30+\left( \frac{7}{2} \right)^2$
$\Rightarrow~ \left(n- \frac{7}{2} \right)^2=30+ \frac{49}{4}$
$\Rightarrow~ \left(n- \frac{7}{2} \right)^2=\frac{120+49}{4}=\frac{169}{4}$
$\Rightarrow ~\left(n- \frac{7}{2} \right)=\frac{\pm 13}{2}$
$\Rightarrow~n=\frac{7}{2}+\frac{13}{2}~~\text{or}~~\frac{7}{2}-\frac{13}{2}$
$\Rightarrow~n=\frac{20}{2}~~\text{or}~~\frac{-6}{2}$
$\Rightarrow~n=10~~\text{or}~~-3$
6. 'n' is a 'number of items'. It cannot be -ve. So we get: n = 10

Part (ii):
1. ⁿP₄ is the permutation of n objects taken 4 at a time.
• Recall that, before learning the factorial notation, we had a lengthy equation:
ⁿP_r = n(n-1)(n-2)(n-3) . . . (n-r+1)
• So we can write:
ⁿP₄ = n(n-1)(n-2)(n-3) . . . (n-4+1)
⇒ ⁿP₄ = n(n-1)(n-2)(n-3)
2. Similarly, ^n-1P₄ = (n-1)(n-2)(n-3) . . . [(n-1)-r+1]
⇒ ^n-1P₄ = (n-1)(n-2)(n-3) . . . [(n-1)-r+1]
⇒ ^n-1P₄ = (n-1)(n-2)(n-3) . . . [n-1-r+1]
⇒ ^n-1P₄ = (n-1)(n-2)(n-3) . . . [n-r]
⇒ ^n-1P₄ = (n-1)(n-2)(n-3) . . . [n-4]
⇒ ^n-1P₄ = (n-1)(n-2)(n-3)(n-4)
3. So the given equation becomes:
$\frac{n(n-1)(n-2)(n-3)}{(n-1)(n-2)(n-3)(n-4)}=\frac{5}{3}$
4. Given that n > 4
• So (n-1), (n-2), (n-3) and (n-4) are all greater than 0.
• So the ratio $\frac{n(n-1)(n-2)(n-3)}{(n-1)(n-2)(n-3)(n-4)}$ exists.
• We can write: $\frac{n(n-1)(n-2)(n-3)}{(n-1)(n-2)(n-3)(n-4)}=\frac{5}{3}$
$\Rightarrow~\frac{n}{(n-4)}=\frac{5}{3}$
$\Rightarrow~3n=5(n-4)$
$\Rightarrow~3n=5n-20$
$\Rightarrow~2n=20$
$\Rightarrow~n=10$

Solved example 7.14
Find the value of r such that,
$5({}^4P_r)=6({}^5P_{r-1})$
Solution:
$\begin{array}{ll}
{}&5({}^4P_r) &{}={}&6({}^5P_{r-1})&{} \\
{\Rightarrow}&5 × \frac{4!}{(4-r)!}&{}={}& 6 × \frac{5!}{[5-(r-1)]!}&{\color {green}{\because ~{}^nP_r=\frac{n!}{(n-r)!}}} \\
{\Rightarrow}&5 × \frac{4!}{(4-r)!}&{}={}& 6 × \frac{5!}{(6-r)!}&{} \\
{\Rightarrow}&5 × \frac{4!}{(4-r)!}&{}={}& 6 × \frac{5 × 4!}{(6-r)!}&{} \\
{\Rightarrow}&\frac{5}{(4-r)!}&{}={}& 6 × \frac{5}{(6-r)!}&{} \\
{\Rightarrow}&\frac{1}{(4-r)!}&{}={}& \frac{6}{(6-r)!}&{} \\
{\Rightarrow}&(6-r)!&{}={}&6(4-r)!&{} \\
{\Rightarrow}&(6-r)(5-r)(4-r)!&{}={}&6(4-r)!&{} \\
{\Rightarrow}&(6-r)(5-r)&{}={}&6&{} \\
{\Rightarrow}&30-6r-5r+r^2&{}={}&6&{} \\
{\Rightarrow}&30-11r+r^2&{}={}&6&{} \\
{\Rightarrow}&r^2-11r&{}={}&-24&{} \\
\end{array}$

• This is a quadratic equation in r. Solving it, we get:
r = 8 or r = 3
• r cannot be greater than the smallest n. In our present case, the smallest n is 4.
• So we can write: r = 3

Solved example 7.15
Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that
(i) all vowels occur together
(ii) all vowels do not occur together.
Solution:
Part (i):
1. In the given word, there are:
    ♦ 3 vowels: A, E and U
    ♦ 5 consonants: D, G, H, T, R
2. The three vowels are to be together always. So we will treat them as one object.
• So now, there is a total of 6 objects:
    ♦ All vowels as one object.
    ♦ All Consonants as different objects.
3. The 6 objects, taken 6 at a time, can be arranged in 6! ways.
• In each of those 6! ways, the three vowels are present together. So in effect, 8 objects are taken together.
4. But in each of those 6! ways, the three vowels can be arranged among themselves in 3! ways.
• So the total number of arrangements = 6! × 3! = 4320
Part (ii):
1. The 8 objects in the word DAUGHTER, can be arranged in 8! ways.
In some of those 8! ways, all the vowels will be together.
In the remaining ways, all the vowels will not be together.
2. So we can write:
Number of arrangements in which all vowels do not occur together =
Total number of arrangements - Number of arrangements in which all vowels occur together
3. Thus we get:
Number of arrangements in which all vowels do not occur together =
[8! - (6! × 3!)] = [8 × 7 × 6! - (6! × 3!)] = [6!(8 × 7 - 3!)] = 36000

Solved example 7.16
In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable ?
Solution:
1. Total number of discs = (4+3+2) = 9
These nine discs can be arranged in 9! ways.
2. Within the 9! ways,
    ♦ The 4 red discs can be arranged among themselves in 4! ways.
        ✰ All those 4! ways will be the same.
    ♦ The 3 yellow discs can be arranged among themselves in 3! ways.
        ✰ All those 3! ways will be the same.
    ♦ The 2 green discs can be arranged among themselves in 2! ways.
        ✰ All those 2! ways will be the same.
3. So the actual number of arrangements = $\frac{9!}{4!\;3!\;2!}=1260$

Solved example 7.17
Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements,
(i) do the words start with P
(ii) do all the vowels always occur together
(iii) do the vowels never occur together
(iv) do the words begin with I and end in P?
Solution:
• In the word INDEPENDENCE, there are 12 letters.
   ♦ N occur 3 times
   ♦ D occur 2 times
   ♦ E occur 4 times
• All other letters occur one time each.
• So the number of arrangements = $\frac{12!}{3!\;2!\;4!}=1663200$

Part (i):
1. There are 12 letters. So consider 12 boxes.
• The first box should be filled with the letter P. It should not be altered.
• So we need to consider the remaining 11 boxes only. Letters in those 11 boxes can be arranged in 11! ways.
2. But N occur 3 times, D occur 2 times, E occur 4 times.
So the number of arrangements = $\frac{11!}{3!\;2!\;4!}=138600$

Part (ii):
1. In the word INDEPENDENCE, vowels are E and I
• There are four Es and one I. So there is a total of 5 vowels.
• These vowels should be treated as one object.
2. There should be one box for the five vowels and seven boxes for the seven consonants.
• The total 8 boxes can be arranged in 8! ways.
3. But there are three Ns and two Ds. So the seven consonants can be arranged in $\frac{8!}{3!\;2!}$ ways.
4. For each of those $\frac{8!}{3!\;2!}$ ways, the five vowels can be arranged in $\frac{5!}{4!}$ ways.
• This is because, there are four Es
5. So the total number of arrangements = $\frac{8!}{3!\;2!}~ × ~\frac{5!}{4!}~=~16800$

Part (iii):
1. Initially, we saw that a total of 1663200 arrangements are possible.
2. In part (ii), we saw that vowels will be together in 16800 arrangements.
3. So we can write:
Vowels will never occur together in (1663200 - 16800) = 1646400 arrangements.

Part (iv):
1. There is only one I and one P.
   ♦ We fill the first box with the letter I.
   ♦ We fill the last box with the letter P.
• Those two boxes should not be altered.
2. The remaining 10 boxes can be arranged in 10! ways.
• But N occur 3 times, D occur 2 times and E occur 4 times.
• So the number of arrangements = $\frac{10!}{3!\;2!\;4!}=12600$

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The link below gives some more solved examples:

Exercise 7.3

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We have completed a discussion on permutations. In the next section, we will see combinations.

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Chapter 7.3 - Number of Permutations Using Factorial Notation

In the previous section, we completed a discussion on factorial notation. In this section, we will see how the factorial notation can help us to shorten the expression for ⁿP_r.

The shortened expression can be obtained in 7 steps:
1. Consider the expression that we derived:
ⁿP_r = n(n-1)(n-2)(n-3) . . . (n-r+1)
• Consider the RHS.
   ♦ n is continuously decreasing. It becomes (n-1), (n-2), (n-3), . . .
   ♦ But it does not become 1.
2. Let us write the remaining terms so that, it becomes 1
   ♦ The last term at present is (n-r+1)
   ♦ So the next term will be (n-r+1-1) = (n-r)
   ♦ The term after (n-r) will be (n-r-1)
   ♦ In this way, we can reach up to 1
3. So the present RHS must be multiplied by [(n-r)(n-r-1) . . . 3 × 2 × 1]
• But we must multiply the denominator also by the same quantity. Otherwise, the value will change.
• So we can write:
${}^nP_r=\frac{n(n-1)(n-2)(n-3)~.~.~.~ (n-r+1)[(n-r)(n-r-1)~.~.~.~ 3 × 2 × 1]}{[(n-r)(n-r-1)~.~.~. 3 × 2 × 1]}$
4. The numerator in the RHS of the expression in (3) is:
$n(n-1)(n-2)(n-3)~.~.~. (n-r+1)(n-r)(n-r-1)~.~.~. 3 × 2 × 1$
• This is n!
5. The denominator in the RHS of the expression in (3) is: $(n-r)(n-r-1)~.~.~. 3 × 2 × 1$
• This is (n-r)!
6. So the expression in (3) becomes: ${}^nP_r=\frac{n!}{(n-r)!}$
7. Now we can write it as the first formula related to permutations.
A. Formula for the number of permutations of n objects taken r at a time is:
${}^nP_r=\frac{n!}{(n-r)!}$
◼ Here two points are to be noted
First point can be written in 4 steps:
(i) n is a number of objects. So it has to be greater than 0
    ♦ That is., 0 < n
(ii) r is a number of objects. So it has to be greater than 0
    ♦ That is., 0 < r
(iii) We are taking r objects out of n objects. So r must be less than or equal to n
    ♦ That is., r ≤ n
(iv) Combining the three inequalities, we get: 0 < r ≤ n
Second point can be written in 3 steps:
(i) The n objects must be different
(ii) For example, suppose that the objects are the letters of the word NUMBER.
Here n is 6. All 6 objects are different.
(iii) Suppose that the objects are the letters of the word ROOT
Here n is 4. All 4 objects are not different because, the letter ‘O’ appears two times.

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Let us see an interesting result based on the above formula. It can be written in 4 steps:
1. We already know this:
If instead of r objects, we take all the n objects, the number of permutations will be:
n(n-1)(n-2)(n-3) . . . 3 × 2 × 1
2. But n(n-1)(n-2)(n-3) . . . 3 × 2 × 1 is n!
• That means, if we take all the n objects, the number of permutations is n!
3. Let us see whether we will get the same result using the formula.
• We have: ${}^nP_r=\frac{n!}{(n-r)!}$
• When r = n, we get:
${}^nP_r={}^nP_n=\frac{n!}{(n-n)!}=\frac{n!}{0!}=\frac{n!}{1}=n!$
• This is the same result that we wrote in (2)
4. So the formula is applicable for r = n also. We can write it as the second formula.
B. Formula for the number of permutations of n objects taken all at a time is: n! 

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Let us see another interesting result based on the above formula. It can be written in 4 steps:
1. We usually take r objects from among the given n objects.
• Some times we take all the n objects.
2. But what if we take none of the objects?
• How many permutations are possible if we take none of the given objects?
The answer can be written in 3 steps:
(i) Number of boxes to be taken:
   ♦ Recall that, for r objects, we take r boxes.
   ♦ For n objects, we take n boxes.
   ♦ So for zero objects, we take zero boxes
So visualize that, there are no boxes to put in the objects.
(ii) There is only one arrangement possible in which no boxes are present.
(iii) That means, if we take zero objects from among the given n objects, the number of permutations possible is 1
3. Let us see whether we will get the same result using the formula.
• We have: ${}^nP_r=\frac{n!}{(n-r)!}$
• When r = 0, we get:
${}^nP_r={}^nP_0=\frac{n!}{(n-0)!}=\frac{n!}{n!}=1$
• This is the same result that we wrote in (2)
4. So the formula is applicable for r = 0 also.

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Now we will derive the third formula.
C. Number of permutations of n objects taken r at a time, when repetition is allowed is: n^r
Proof can be written in 2 steps:
1.If r objects are taken, there will be r boxes
• The first box can be filled in n different ways.
• Since repetition is allowed, the second box can also be filled in n different ways.
• Since repetition is allowed, the third box can also be filled in n different ways.
- - -
- - -
• so on . . .
• This can be visualized as shown in fig.7.6 below:

Fig.7.6

2. So by applying the multiplication principle, the number of permutations possible = (n × n × n . . . n × n) = n^r
• Note that, if all the n objects are taken, we will get nⁿ

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Before we move on to the next formula, let us apply the above formulas to the examples that we saw in earlier sections
Example 1:
We want to find the number of permutations when the letters of the word ROSE are taken all at a time. Repetition not allowed.
Solution:
1. We must use formula B because all objects are taken at a time.
So we get: Number of permutations = n! = 4! = 4 × 3 × 2 × 1=24
2. When repetition is allowed, we apply formula C: n^r
• So we get: $n^r = 4^4 = 256$

Example 2:
We want to find the number of permutations when the letters of the word NUMBER are taken 3 at a time. Repetition not allowed.
Solution:
1. We have formula A: ${}^nP_r=\frac{n!}{(n-r)!}$
   ♦ Here n = 6 and r = 3
• So we get:
${}^nP_r={}^6P_3=\frac{6!}{(6-3)!}=\frac{6!}{3!}=\frac{6 × 5 × 4 × 3!}{3!}=6 × 5 × 4=120$
2. When repetition is allowed, we apply formula C: n^r
• So we get: $n^r = 6^3 = 216$

Another example:
In a group, there are 12 members. A Chairman and a Vice-Chairman are to be selected from among them. In how many different ways can this be done. None of the members can hold more than one position.
Solution:
We have formula A: ${}^nP_r=\frac{n!}{(n-r)!}$
   ♦ Here n = 12 and r = 2
• So we get:
${}^nP_r={}^{12}P_2=\frac{12!}{(12-2)!}=\frac{12!}{10!}=\frac{12 × 11 × 10!}{10!}=12 × 11=132$

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The next formula that we are going to derive is applicable when some of the objects are not different. Some basics can be written in steps:
1. The formulas that we obtained so far were related to those cases, where all objects are different.
• For example,
   ♦ All letters of the word ROSE are different.
   ♦ All letters of the word NUMBER are different.
2. Now consider the word ROOT. We want to calculate the number of permutations of the letters in this word.
• In this word, there are 4 objects.  But two of them (the two Os) are the same.
3. For convenience, let us assume that the two identical objects are different.
• So we will name one of the Os as O₁ and the other as O₂. Now there are 4 different objects.
4. We want the number of permutations of those 4 objects, taking all at a time. Repetition is not allowed.
• We can use formula B: n!
   ♦ Here n = 4
• So we get:
Number of permutations = n! = 4! = 4 × 3 × 2 × 1 = 24
5. We can write:
If we make a list of all permutations, there will be 24 permutations in that list.
• Some of those permutations are:
   ♦ RO₁O₂T
   ♦ RO₂O₁T
   ♦ O₁TO₂R
   ♦ O₂TO₁R, etc.,
6. Now consider the set of two permutations: RO₁O₂T and RO₂O₁T  
• The O₁ and O₂ are same as O
• That means, the two permutations RO₁O₂T and RO₂O₁T will reduce to one permutation: ROOT.
7. Let us see another set of two permutations: O₁TO₂R and O₂TO₁R
• The O₁ and O₂ are same as O
• That means, the two permutations O₁TO₂R and O₂TO₁R reduce to OTOR.
8. In this way, if we discard the subscripts ‘1’ and ‘2’,
   ♦ for every permutation, there will be an identical permutation.
• We can write:
[No. of actual permutations] × 2 = 24
• So [No. of actual permutations] = $\frac{24}{2}$ = 12
9. In the above step,
   ♦ [No. of actual permutations] is multiplied by ‘2’.
• The ‘2’ is actually 2!
   ♦ This is because, the two Os can be arranged among themselves in 2! ways.
10. Thus we can write:
Number of permutations when all letters of the word ROOT are taken is: 12

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Let us see another example. It can be written in steps:
1. Consider the word INSTITUTE
• There are 9 letters.
   ♦ I is present 2 times
   ♦ T is present 3 times
   ♦ All remaining letters are different.
• So the number of permutations when all letters of the word INSTITUTE are taken is: $\frac{9!}{2!\;3!}$

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Based on the above examples of ROOT and INSTITUTE, we can write the fourth formula.
D. Number of permutations of n objects when p objects among them are of the same kind and the rest are all different is: $\frac{n!}{p!}$
• This formula is applicable when there is only one type of recurring object. That recurring object will be present p times.

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• Some times there will be more than one type of recurring objects.
   ♦ The first type will be present p₁ times.
   ♦ The second type will be present p₂ times.
   ♦ The third type will be present p3q times.
   ♦ - - -
   ♦ - - -
   ♦ The k^th type will be present p_k times.
• In such cases, we can use the fifth formula:
E. Number of permutations of n objects when there are more than one type of recurring objects is: $\frac{n!}{{p_1}!\, {p_2}!\, {p_3}! ~.~.~.~{p_k}!}$

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Now we will see a solved example:

Solved example 7.10
Find the number of permutations of the letters of the word ALLAHABAD.
Solution:
• There are 9 letters.
   ♦ A is present 4 times
   ♦ L is present 2 times
   ♦ All remaining letters are different.
• So the number of permutations =
$\frac{9!}{4!\;2!}=~\frac{9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1}{4 × 3 × 2 × 1 × 2 × 1}=\frac{9 × 8 × 7 × 6 × 5}{2 × 1}=9 × 8 × 7 × 3 × 5 = 7560$

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In the next section, we will see a few more solved examples.

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Chapter 7.1 - Application of Multiplication Principle

In the previous section, we saw the multiplication principle. We also saw some solved examples. In this section, we will see a few more solved examples.

Solved example 7.3
How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?
Solution:
1. Fig.7.4(a) below shows a possible combination of two numbers from the given list. The numbers are placed side by side.

Fig.7.4

• Fig.(b) shows another possible combination.
• Note that in both figs. (a) and (b), the two digit number formed, is even. This is because, the digit in the units place is even.
◼ We have to find the answer to this question:
How many such combinations are possible.
2. To try the various possible combinations, we can take two boxes as shown in fig.c.
• We have to fill the boxes with the given five digits, taking two at a time.
3. First we will fill the left side box.
• So filling of the left side box is the first event. It can take place in five different ways.
4. Next we fill the right side box. So filling the right side box is the second event.
• After filling the left box, we will be having four digits.
• But since repetition of digits is allowed, we have the option to use all the five digits for filling the right side box.
• But again, the two digit number formed must be even. So out of the five given digits, we have the option to use only two. They are: 2 and 4
• So we can write:
The second event can take place in two different ways.
5. So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 2) = 10

Solved example 7.4
Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available.
Solution:
• Five different flags are available.
◼ Note the specification in the question: at least two flags
• That means, we can make signals by
   ♦ putting 2 flags together      
   ♦ putting 3 flags together      
   ♦ putting 4 flags together      
   ♦ putting 5 flags together
• Making a signal using a single flag is not allowed. That is why ‘at least two flags’ is specified.
1. To try various combinations using 2 flags, we can use the two boxes shown in fig.7.5(a) below:

Fig.7.5

   ♦ The top box can be filled in 5 different ways.
   ♦ The bottom box can be filled in 4 different ways
• So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 4) = 20    
2. To try various combinations using 3 flags, we can use the three boxes shown in fig.7.5(b) above.
   ♦ The top box can be filled in 5 different ways.
   ♦ The box below it can be filled in 4 different ways
   ♦ The bottom box can be filled in 3 different ways
• So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 4 × 3) = 60
3. To try various combinations using 4 flags, we can use the four boxes shown in fig.7.5(c) above.
   ♦ The top box can be filled in 5 different ways.
   ♦ The box below it can be filled in 4 different ways
   ♦ The box below it can be filled in 3 different ways
   ♦ The bottom box can be filled in 2 different ways
• So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 4 × 3 × 2) = 120
3. To try various combinations using 5 flags, we can use the five boxes shown in fig.7.5(d) above.
   ♦ The top box can be filled in 5 different ways.
   ♦ The box below it can be filled in 4 different ways
   ♦ The box below it can be filled in 3 different ways
   ♦ The box below it can be filled in 2 different ways
   ♦ The bottom box can be filled in 1 way
• So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 4 × 3 × 2 × 1) = 120
4. Now we can add all the possible combinations. We get:
20 + 60 + 120 + 120 = 320

Solved example 7.5
How many four digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5 if repetition of digits is not allowed?
Solution:
1. To try various combinations, we can place four boxes side by side.
2. We are given 6 digits: 0, 1, 2, 3, 4, and 5. But we cannot use ‘0’ in the first box. Because, then the number formed would be a three digit number.
• So we have 5 digits to try on the first box. That means, the first box can be filled in 5 different ways.
3. There are 6 digits available to fill in the second box. But since repetition is not allowed, we can use only 5 digits. That means, the second box can be filled in 5 different ways.
4. In a similar way,
• the third box can be filled in 4 different ways.
• the fourth box can be filled in 3 different ways.
5. So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 5 × 4 × 3) = 300

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The link below gives some more solved examples:

Exercise 7.1

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• In the next section, we will see permutations.

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Chapter 7 - Permutations And Combinations

In the previous section, we completed a discussion on linear inequalities. In this chapter we will see permutations and combinations.

We can obtain a basic understanding about permutations and combinations by considering an example. It can be written in 4 steps:
1. Suppose that, we have a number lock.
• Let it have  four wheels.
• Let each wheel have seven digits from 0 to 6
(Some images can be seen here)
2. We will be able to open the lock only if we can remember a secret combination of four digits.
• We must bring those four digits in a particular line.
• Also, those four digits must be arranged in the correct sequence.
For example, if the secret combination is 4, 2, 3, 6:
‘4’ must always be in the first position.  
‘2’ must always be in the second position.  
‘3’ must always be in the third position.  
‘6’ must always be in the fourth position.  
Interchanging of positions is not allowed.
• So it is important to remember all the four digits. It is also important to remember the correct positions.
3. Suppose that, we have forgotten the digits and the positions. Then the only way to open the lock is to try various combinations. Let us analyze such a situation:
• We remember the first digit. Let it be ‘4’
• Also we remember that, no digit repeats.
• Then some of the possible combinations are:
4, 6, 3, 1  
4, 3, 5, 2  
4, 1, 6, 3  
4, 2, 5, 6  
so on . . .
4. Obviously, it is a tedious process to obtain the correct combination by doing such trials. We will have to make a very large number of trials.
• Two questions will arise in our minds:
    ♦ How many trial combinations are possible?
    ♦ Is there a systematic way to carry out the trials?
• The principles of permutations and combinations help us to answer those questions.

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• First we will learn about the Fundamental principle of counting
• This principle can be explained using two examples.
Example 1:
This can be written in 3 steps:
1. A student has two pairs of pants. One is blue and the other is brown.
• Also he has three shirts. Green, yellow and red.
◼ In how many different ways can he dress up?
2. This problem can be analyzed effectively with the help of fig.7.1 below.
It can be written in three steps.

Fig.7.1

(i) Suppose that the student selects blue pants.
• Taking that blue pants, he can move along the arrow no. 1 to select the green shirt.
• Taking that blue pants, he can move along the arrow no. 2 to select the yellow shirt.
• Taking that blue pants, he can move along the arrow no. 3 to select the red shirt.
• So it is clear that, if the blue pants is selected, there are three different ways in which he can dress up.  
(ii) Suppose that the student selects brown pants.
• Taking that brown pants, he can move along the arrow no. 4 to select the green shirt.
• Taking that brown pants, he can move along the arrow no. 5 to select the yellow shirt.
• Taking that brown pants, he can move along the arrow no. 6 to select the red shirt.
• So it is clear that, if the brown pants is selected, there are three different ways in which he can dress up.
(iii) Thus altogether, there are six different ways in which he can dress up.
3. We obtained an important result:
The total number of possible combinations is 6
• This result can be obtained with out writing all the above detailed steps. There is an easy method. It can be written in three steps:
(i) The first event is: Selecting a pair of pants.
• This event can occur in two different ways.
(ii) The second event is: Selecting a shirt.
• For each of the two ways for the first event, the second event can occur in three different ways.
(iii) So the total number of combinations possible is (2 × 3) = 6

Example 2:
This can be written in 3 steps:
1. A student has two pairs of pants. One is blue and the other is brown.
• Also he has three shirts. Green, yellow and red.
• Also he has two bags. Orange and magenta.
◼ In how many different ways can he dress up?
2. This problem can be analyzed effectively with the help of fig.7.2 below.
It can be written in three steps.

Fig.7.2

(i) Consider the dashed line in fig.7.2.
• The left side of the dashed line is the same fig. that we saw in example 1
• That means, the dashed line indicates the six possible combinations of the two pants and three shirts.
(ii) After dressing up in any one of those six combinations, the student can select either the orange bag or the magenta bag.
• That means, for each of the final six combinations that we saw in the previous example 1, there are two bags.
(iii) Thus altogether, there will be twelve different ways in which he can dress up.
3. We obtained an important result:
The total number of possible combinations is 12
• This result can be obtained with out writing all the above detailed steps. There is an easy method. It can be written in three steps:
(i) The first event is: Selecting a pair of pants.
• This event can occur in two different ways.
(ii) The second event is: Selecting a shirt.
• For each of the two ways for the first event, the second event can occur in three different ways.
(iii) So the total number of combinations possible from the first and second events is (2 × 3)
(iv) The third event is: Selecting a bag.
• For each of the (2 × 3) combinations of the first and second events, the third event can occur in two different ways.
(iii) So the total number of combinations possible from the first, second and third events is (2 × 3) × 2 = (2 × 3 × 2) = 12

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Based on the above two examples, we can write the fundamental principle of counting:
A. When there are two events.
This can be written in 3 steps:
(i) The first event can occur in m different ways.
(ii) For each of the m ways, the second event can occur in n different ways.
(iii) Then the total number of combinations possible is (m ×  n)
B. For three events:
This can be written in steps:
(i) The first event can occur in m different ways.
(ii) For each of the m ways, the second event can occur in n different ways.
(iii)Then the total number of combinations possible from the first and second events is (m × n)
(iv) For each of the (m × n) ways, the third event can occur in p different ways.
(v) Then the total number of combinations possible from the three events is (m × n × p)

◼ In this way, the principle can be written for any number of finite events.

◼ The fundamental principle of counting is also known as multiplication principle.

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Let us see a solved example:

Solved example 7.1
Find the number of 4 letter words, with or without meaning, which can be
formed out of the letters of the word ROSE, where the repetition of the letters is not allowed.
Solution:
1. Consider a four letter word.
• The first letter of that word can be R, O, S or E
• The first event is ‘selecting the first letter’.
• It is clear that, the first event can occur in four different ways.
2. The second event is ‘selecting the second letter’.
• Remember that no repetition is allowed. So we can write:
    ♦ If R is selected in the first event, the letter for the second event can be only O, S or E
    ♦ If O is selected in the first event, the letter for the second event can be only R, S or E
    ♦ If S is selected in the first event, the letter for the second event can be only R, O or E
    ♦ If E is selected in the first event, the letter for the second event can be only R, O or S
• It is clear that, for each letter selected in the first event, the second event can occur in three different ways.
3. The third event is ‘selecting the third letter’.
• Remember that no repetition is allowed. So it is clear that, the third event can occur in two different ways.  
4. The fourth event is ‘selecting the fourth letter’.
• Remember that no repetition is allowed. So it is clear that, the fourth event can occur in one way only.
5. So by applying multiplication principle, we get:
Total number of combinations possible = (4 × 3 × 2 × 1) = 24

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• In the above problem, what will happen if repetition is allowed?
The answer can be written in two steps:
1. If repetition is allowed,
    ♦ The first event can occur in four different ways.       
    ♦ The second event can occur in four different ways.       
    ♦ The third event can occur in four different ways.       
    ♦ The fourth event can occur in four different ways.       
2. So the total number of combinations possible = (4 × 4 × 4 × 4) = 256 

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Solved example 7.2
We are given 4 flags of different colors. How many different signals can be generated, if a signal requires the use of 2 flags. One placed above and the other below?
Solution:
1. Fig.7.3(a) shows a combination of two flags. One placed above and the other below.

 

• Fig.(b) shows another combination. We have to find out the maximum possible number of such combinations.
2. To do the trials, we can take two boxes as shown in fig.c.
• We have to fill the boxes with the given four flags, taking two at a time.
3. First we will fill the top box.
• So filling of the top box is the first event. It can take place in four different ways.
4. Next we fill the bottom box. So filling the bottom box is the second event.
• After filling the top box, we will be having three flags. So the second event can take place in three different ways.
5. So by applying multiplication principle, we get:
Total number of combinations possible = (4 × 3) = 12

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• In the next section, we will see a few more solved examples.

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