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:
nPr = 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:
nPr = 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 = $\frac{10!}{3!\;2!\;4!}=12600$
The link below gives some more solved examples:
We have completed a discussion on permutations. In the next section, we will see combinations.
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