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 7.2 - Permutations

In the previous section, we completed a discussion on the multiplication principle. In this section, we will see Permutations.

Some basics about permutations can be understood by considering the example that we saw in a previous section. It can be written in 3 steps:
1. In the example 7.1 in the first section of this chapter, we arranged the four letters of the word ROSE in different possible ways.
• We saw that, the letters can be arranged in 24 different ways.
   ♦ So we have a list of 24 words.
   ♦ Each of the 24 words in that list is unique.
2. This uniqueness can be explained in 3 steps:
(i) Let us consider one of the word in the list, say SREO
(ii) In this word,
   ♦ S is in the first position
   ♦ R is in the second position
   ♦ E is in the third position
   ♦ O is in the fourth position
• We must not change this order.
(iii) If we change the order, we will get a ‘different word’.
   ♦ This ‘different word’ will be already present in the list.
   ♦ Then we cannot say that, the list contains 24 different words.
   ♦ Because one of them is repeating.
   ♦ The list will then become invalid.
• That is why, we say that, each member of the list is unique. We must not alter the arrangement in that member.
3. Each arrangement is called a permutation of 4 different letters taken all at a time.
◼ We can write:
There are 24 permutations.

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Let us see another example. It can be written in 3 steps:
1. Consider the word NUMBER.
How many 3 letter words, with or without meaning, can be formed using the letters in this word. Repetition of letters is not allowed.
Answer: There are 6 letters in the word NUMBER. So applying the multiplication principle, the answer is (6 × 5 × 4) = 120
   ♦ So we have a list of 120 words.
         ✰ Some examples are: NBR, MRU, RNE etc.,
   ♦ Each of the 120 words in that list is unique.
2. This uniqueness can be explained in 3 steps:
(i) Let us consider one of the word in the list, say RNE
(ii) In this word,
   ♦ R is in the first position
   ♦ N is in the second position
   ♦ E is in the third position
• We must not change this order.
(iii) If we change the order, we will get a ‘different word’.
   ♦ This ‘different word’ will be already present in the list.
   ♦ Then we cannot say that, the list contains 120 different words.
   ♦ Because one of them is repeating.
   ♦ The list will then become invalid.
• That is why, we say that, each member of the list is unique. We must not alter the arrangement in that member.
3. Each arrangement is called a permutation of 6 different letters taken 3 at a time.
◼ We can write:
There are 120 permutations.

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Now we can write the definition for permutation. It can be written in 3 steps:
(i) A permutation is an arrangement of objects in a definite order.
(ii) In some problems, all available objects will be included in each of the arrangements.
(iii) In some problems, only a certain number of the available objects will be included in each of the arrangements.

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Next we will try to derive formulas for various types of problems.
A. Formula for the number of permutations of n objects taken r at a time
This formula can be derived in 3 steps:
1. Suppose that there are a total of n objects.
• If we take r objects at a time and try the permutations, then the number of permutations possible is given by: n(n-1)(n-2)(n-3) . . . (n-r+1)
2. The proof can be written in 2 steps:
(i) If we take r objects at a time, there will be r boxes
• The first box can be filled in n different ways
• The second box can be filled in (n-1) different ways
    ♦ (n-1) is [n-(2-1)]
    ♦ So for second box, we have the term (2-1)
• The third box can be filled in (n-2) different ways
    ♦ (n-2) is [n-(3-1)]
    ♦ So for third box, we have the term (3-1)
• The fourth box can be filled in (n-3) different ways
    ♦ (n-3) is [n-(4-1)]
    ♦ So for fourth box, we have the term (4-1)
- - -
- - -
• Based on the pattern, we get:
The r^th box can be filled in [n-(r-1)] different ways
⇒ The r^th box can be filled in (n-r+1) different ways.
(ii) So by applying the multiplication principle, we get:
Total number of arrangements (permutations) = n(n-1)(n-2)(n-3) . . . (n-r+1)
3. The expression n(n-1)(n-2)(n-3) . . . (n-r+1) is denoted as ⁿP_r
◼ So we can write the formula:
The number of permutations of n objects taken r at a time is given by:
ⁿP_r = n(n-1)(n-2)(n-3) . . . (n-r+1)
◼ 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 ROOTS
Here n is 5. All 5 objects are not different because, the letter ‘O’ appears two times.

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• So we obtained an expression for calculating the number of permutations:
ⁿP_r = n(n-1)(n-2)(n-3) . . . (n-r+1)
• This expression is cumbersome. We must try to shorten it.
• The symbol n! can help us to shorten the expression.
• So our next task is to learn about n!. It can be written in steps:
1. n! is read as factorial n. It can also be read as n factorial.
2. When we write n!, the n must be a natural number.
3. n! is the product of all natural numbers beginning from 1 and ending at n
• So we can write:
n! = 1 × 2 × 3 × 4 × . . .  × (n-1) × n
• Let us see some examples:
1! = 1
2! = 2 × 1
3! = 3 × 2 × 1
4! = 4 × 3 × 2 × 1
5! = 5 × 4 × 3 × 2 × 1
- - -
- - -
so on . . .
4. Mathematicians define 0! as 1
    ♦ That is., 0! = 1
• We will see the proof in later sections.
5. Now we will see some interesting results. It can be written in 4 steps:
(i) We know that, 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
• This can be written as: 7! = 7 × [6 × 5 × 4 × 3 × 2 × 1]
    ♦ But [6 × 5 × 4 × 3 × 2 × 1] is 6!
• So we get: 7! = 7 × 6!
• Based on this example, we can write: n!= n × (n-1)!
(ii) We know that, 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
• This can be written as: 7! = 7 × 6 × [5 × 4 × 3 × 2 × 1]
    ♦ But [5 × 4 × 3 × 2 × 1] is 5!
• So we get: 7! = 7 × 6 × 5!
• Based on this example, we can write: n!= n × (n-1) × (n-2)!
(Here n must be greater than or equal to 2. Otherwise, (n-2) will become -ve.)
(iii) We know that, 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
• This can be written as: 7! = 7 × 6 × 5 × [4 × 3 × 2 × 1]
    ♦ But [4 × 3 × 2 × 1] is 4!
• So we get: 7! = 7 × 6 × 5 × 4!
• Based on this example, we can write: n!= n × (n-1) × (n-2) × (n-3)!
(Here n must be greater than or equal to 3. Otherwise, (n-3) will become -ve.)
(iv) So we can write a general form:
n!= n × (n-1)!
n!= n × (n-1) × (n-2)!  [provided n ≥ 2]
n!= n × (n-1) × (n-2) × (n-3)!  [provided n ≥ 3]
- - -
- - -
so on . . .

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Now we will see some solved examples

Solved example 7.6
Evaluate (i) 5!  (ii) 7!  (iii) 7! - 5!
Solution:
Part (i):
5! = 5 × 4 × 3 × 2 × 1 = 120

Part (ii):
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
Alternate method:
7! = 7 × 6 × 5! = 42 × 120 = 5040

Part (iii):
7! - 5! = (5040 - 120) = 4920

Solved example 7.7
Evaluate (i) $\frac{7!}{5!}$  (ii) $\frac{12!}{(10!)(2!)}$
Solution:
Part (i):
$\begin{array}{ll}
{}&\frac{7!}{5!} &{}={}& \frac{7 × 6 × 5!}{5!} &{} \\
{}&\phantom{\frac{7!}{5!}}&{}={}&7 × 6 \\
{}&\phantom{\frac{7!}{5!}}&{}={}&42 \\
\end{array}$

Part (ii):
$\begin{array}{ll}
{}&\frac{12!}{(10!)(2!)} &{}={}& \frac{12 × 11 × 10!}{(10!)(2 × 1)} &{} \\
{}&\phantom{\frac{12!}{(10!)(2!)}}&{}={}&\frac{12 × 11}{(2 × 1)} \\
{}&\phantom{\frac{12!}{(10!)(2!)}}&{}={}&6 × 11 \\
{}&\phantom{\frac{12!}{(10!)(2!)}}&{}={}&66 \\
\end{array}$

Solved example 7.8
Evaluate $\frac{n!}{r! (n-r)!}$ when n = 5, r = 2
Solution:
$\begin{array}{ll}
{}&\frac{n!}{r! (n-r)!} &{}={}& \frac{5!}{2! (5-2)!}\\
{}&\phantom{\frac{n!}{r! (n-r)!}}&{}={}&\frac{5 × 4 × 3!}{2! (3)!} \\
{}&\phantom{\frac{n!}{r! (n-r)!}}&{}={}&\frac{5 × 4}{2 × 1} \\
{}&\phantom{\frac{n!}{r! (n-r)!}}&{}={}&5 × 2 \\
{}&\phantom{\frac{n!}{r! (n-r)!}}&{}={}&10 \\
\end{array}$

Solved example 7.9
$\rm{If}~\frac{1}{8!}+\frac{1}{9!}=\frac{x}{10!},~\rm{find x}$
Solution:
$\begin{array}{ll}
{}&\frac{1}{8!}+\frac{1}{9!} &{}={}& \frac{x}{10!}&{} \\
{\Rightarrow}&\frac{1}{8!}+\frac{1}{9!}&{}={}& \frac{x}{10 × 9 × 8!}&{} \\
{\Rightarrow}&\frac{10 × 9 × 8!}{8!}+\frac{10 × 9 × 8!}{9!}&{}={}& \frac{x × 10 × 9 × 8!}{10 × 9 × 8!}&{\color {green}{\text{(Multiplying by 10 × 9 × 8!)}}} \\
{\Rightarrow}&\frac{10 × 9 × 8!}{8!}+\frac{10 × 9 × 8!}{9 × 8!}&{}={}& \frac{x × 10 × 9 × 8!}{10 × 9 × 8!}&{} \\
{\Rightarrow}&10 × 9+10&{}={}& x&{} \\
{\Rightarrow}&90+10&{}={}& x&{} \\
{\Rightarrow}&100&{}={}& x&{} \\
\end{array}$

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The link below gives some more solved examples:

Exercise 7.2

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Now we have a basic idea about factorial notation. Remember that, we are trying to shorten the expression for ⁿP_r. In the next section, we will see how the factorial notation will help us to achieve this.

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