Examples (Revised) - Chapter 6 Permutations & Combinations class 11 ncert solutions Maths - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Chapter 6: Permutations & Combinations - NCERT Solutions Class 11 Maths

Example 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

There are as many words as there are ways of filling in 4 vacant places
   by the 4 letters, keeping in mind that the repetition is not allowed. The first place can be filled in 4 different ways by anyone of the 4 letters R,O,S,E. Following which, the second place can be filled in by anyone of the remaining 3 letters in 3 different ways, following which the third place can be filled in 2 different ways; following which, the fourth place can be filled in 1 way. Thus, the number of ways in which the 4 places can be filled, by the multiplication principle, is $4 \times 3 \times 2 \times 1=24$. Hence, the required number of words is 24 .

Example 2

Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other?
Solution

There will be as many signals as there are ways of filling in 2 vacant places
in succession by the 4 flags of different colours. The upper vacant place can be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant place can be filled in 3 different ways by anyone of the remaining 3 different flags. Hence, by the multiplication principle, the required number of signals $=4 \times 3=12$.
Example 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

There will be as many ways as there are ways of filling 2 vacant places
  in succession by the five given digits. Here, in this case, we start filling in unit's place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten's place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is $2 \times 5$, i.e., 10 .
Example 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

A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us count the possible number of signals consisting of 2 flags, 3 flags, 4 flags and 5 flags separately and then add the respective numbers.
There will be as many 2 flag signals as there are ways of filling in 2 vacant places

in succession by the 5 flags available. By Multiplication rule, the number of ways is $5 \times 4=20$.

Similarly, there will be as many 3 flag signals as there are ways of filling in 3  in succession by the 5 flags.

The number of ways is $5 \times 4 \times 3=60$.
Continuing the same way, we find that
The number of 4 flag signals $=5 \times 4 \times 3 \times 2=120$
and the number of 5 flag signals $=5 \times 4 \times 3 \times 2 \times 1=120$
Therefore, the required no of signals $=20+60+120+120=320$.

Example 5

Evaluate (i) 5 !
(ii) 7 !
(iii) 7 ! -5 !

Solution
(i) 5 ! $=1 \times 2 \times 3 \times 4 \times 5=120$
(ii) $7!=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7=5040$
and
(iii) $7!-5!=5040-120=4920$.

Example 6

Compute (i) $\frac{7!}{5!}$
(ii) $\frac{12!}{(10!)(2!)}$

Solution
(i) We have $\frac{7!}{5!}=\frac{7 \times 6 \times 5!}{5!}=7 \times 6=42$
and
(ii) $\frac{12!}{(10!)(2!)}=\frac{12 \times 11 \times(10!)}{(10!) \times(2)}=6 \times 11=66$.

Example 7

Evaluate $\frac{n!}{r!(n-r)!}$, when $n=5, r=2$.
Solution

We have to evaluate $\frac{5!}{2!(5-2)!}$ (since $\left.n=5, r=2\right)$
We have $\quad \frac{5!}{2!(5-2)!}=\frac{5!}{2!\times 3!}=\frac{5 \times 4}{2}=10$.
Example   8

$\quad$ If $\frac{1}{8!}+\frac{1}{9!}=\frac{x}{10!}$, find $x$.
Solution

We have $\frac{1}{8!}+\frac{1}{9 \times 8!}=\frac{x}{10 \times 9 \times 8!}$

Therefore $1+\frac{1}{9}=\frac{x}{10 \times 9}$ or $\frac{10}{9}=\frac{x}{10 \times 9}$ So $\quad x=100$.

Example 9

Find the number of permutations of the letters of the word ALLAHABAD.

Solution

Here, there are 9 objects (letters) of which there are 4A's, 2 L's and rest are all different.
Therefore, the required number of arrangements $=\frac{9!}{4!2!}=\frac{5 \times 6 \times 7 \times 8 \times 9}{2}=7560$
Example 10

How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?

Solution

Here order matters for example 1234 and 1324 are two different numbers. Therefore, there will be as many 4 digit numbers as there are permutations of 9 different digits taken 4 at a time.
Therefore, the required 4 digit numbers $={ }^9 \mathrm{P}_4=\frac{9!}{(9-4)!}=\frac{9!}{5!}=9 \times 8 \times 7 \times 6=3024$.
Example 11

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

Every number between 100 and 1000 is a 3-digit number. We, first, have to

count the permutations of 6 digits taken 3 at a time. This number would be ${ }^6 P_3$. But, these permutations will include those also where 0 is at the 100 's place. For example, $092,042, \ldots$, etc are such numbers which are actually 2-digit numbers and hence the number of such numbers has to be subtracted from ${ }^6 \mathrm{P}_3$ to get the required number. To get the number of such numbers, we fix 0 at the 100 's place and rearrange the remaining 5 digits taking 2 at a time. This number is ${ }^5 \mathrm{P}_2$. So
The required number $\quad={ }^6 \mathrm{P}_3-{ }^5 \mathrm{P}_2=\frac{6!}{3!}-\frac{5!}{3!}$
$
=4 \times 5 \times 6-4 \times 5=100
$

Example 12

Find the value of $n$ such that
(i) ${ }^n \mathrm{P}_5=42{ }^n \mathrm{P}_3, n>4$
(ii) $\frac{{ }^n \mathrm{P}_4}{{ }^{n-1} \mathrm{P}_4}=\frac{5}{3}, n>4$

Solution

(i) Given that
$
{ }^n \mathrm{P}_5=42{ }^n \mathrm{P}_3
$
or
$
n(n-1)(n-2)(n-3)(n-4)=42 n(n-1)(n-2)
$

Since
$
n>4
$
so $n(n-1)(n-2) \neq 0$
Therefore, by dividing both sides by $n(n-1)(n-2)$, we get
$
\begin{array}{ll} 
& (n-3(n-4)=42 \\
\text { or } & n^2-7 n-30=0 \\
\text { or } & n^2-10 n+3 n-30 \\
\text { or } & (n-10)(n+3)=0 \\
\text { or } & n-10=0 \text { or } \quad n+3=0 \\
\text { or } & n=10 \quad \text { or } \quad n=-3
\end{array}
$

As $n$ cannot be negative, so $n=10$.
(ii) Given that $\frac{{ }^n \mathrm{P}_4}{{ }^{n-1} \mathrm{P}_4}=\frac{5}{3}$

Therefore $3 n(n-1)(n-2)(n-3)=5(n-1)(n-2)(n-3)(n-4)$
or
$3 n=5(n-4) \quad[$ as $(n-1)(n-2)(n-3) \neq 0, n>4]$
or $n=10$.

Example 13

Find $r$, if $5{ }^4 \mathrm{P}_r=6{ }^5 \mathrm{P}_{r-1}$.
Solution

We have $5{ }^4 \mathrm{P}_r=6{ }^5 \mathrm{P}_{r-1}$
or
$
5 \times \frac{4!}{(4-r)!}=6 \times \frac{5!}{(5-r+1)!}
$
or
$
\frac{5!}{(4-r)!}=\frac{6 \times 5!}{(5-r+1)(5-r)(5-r-1)!}
$
or
$
(6-r)(5-r)=6
$
or
$
r^2-11 r+24=0
$
or
$
r^2-8 r-3 r+24=0
$
or
$
(r-8)(r-3)=0
$
or
$
r=8 \text { or } r=3 \text {. }
$

Hence $\quad r=8,3$.
Example 14

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

(i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be ${ }^6 \mathrm{P}_6=6$ !. Corresponding to each of these permutations, we shall have 3 ! permutations of the three vowels $\mathrm{A}, \mathrm{U}, \mathrm{E}$ taken all at a time . Hence, by the multiplication principle the required number of permutations $=6!\times 3!=4320$.
(ii) If we have to count those permutations in which all vowels are never together, we first have to find all possible arrangments of 8 letters taken all at a time, which can be done in 8 ! ways. Then, we have to subtract from this number, the number

of permutations in which the vowels are always together.
Therefore, the required number
$
\begin{aligned}
8!-6!\times 3! & =6!(7 \times 8-6) \\
& =2 \times 6!(28-3) \\
& =50 \times 6!=50 \times 720=36000
\end{aligned}
$

Example 15

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

Total number of discs are $4+3+2=9$. Out of 9 discs, 4 are of the first kind 

(red), 3 are of the second kind (yellow) and 2 are of the third kind (green).
Therefore, the number of arrangements $\frac{9!}{4!3!2!}=1260$.
Example 16

Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements,
(i) do the words start with $\mathrm{P}$
(ii) do all the vowels always occur together
(iii) do the vowels never occur together
(iv) do the words begin with $\mathrm{I}$ and end in $\mathrm{P}$ ?

Solution

There are 12 letters, of which $\mathrm{N}$ appears 3 times, $\mathrm{E}$ appears 4 times and $\mathrm{D}$ appears 2 times and the rest are all different. Therefore
The required number of arrangements $=\frac{12!}{3!4!2!}=1663200$
(i) Let us fix $\mathrm{P}$ at the extreme left position, we, then, count the arrangements of the remaining 11 letters. Therefore, the required number of words starting with $\mathrm{P}$
$
=\frac{11!}{3!2!4!}=138600 \text {. }
$
(ii) There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have to always occur together, we treat them as a single object EEEEI for the time being. This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are $3 \mathrm{Ns}$ and $2 \mathrm{Ds}$, can be rearranged in
$\frac{8!}{3!2!}$ ways. Corresponding to each of these arrangements, the 5 vowels E, E, E,

E and I can be rearranged in $\frac{5!}{4!}$ ways. Therefore, by multiplication principle, the required number of arrangements
$
=\frac{8!}{3!2!} \times \frac{5!}{4!}=16800
$
(iii) The required number of arrangements $=$ the total number of arrangements (without any restriction) - the number of arrangements where all the vowels occur together.

$
=1663200-16800=1646400
$
(iv) Let us fix $\mathrm{I}$ and $\mathrm{P}$ at the extreme ends ( $\mathrm{I}$ at the left end and $\mathrm{P}$ at the right end).
We are left with 10 letters.
Hence, the required number of arrangements
$
=\frac{10!}{3!2!4!}=12600
$

Example 17

If ${ }^n \mathrm{C}_9={ }^n \mathrm{C}_8$, find ${ }^n \mathrm{C}_{17}$.
Solution

We have ${ }^n \mathrm{C}_9={ }^n \mathrm{C}_8$
i.e., $\quad \frac{n!}{9!(n-9)!}=\frac{n!}{(n-8)!8!}$
or $\quad \frac{1}{9}=\frac{1}{n-8}$ or $n-8=9$ or $n=17$
Therefore ${ }^n \mathrm{C}_{17}={ }^{17} \mathrm{C}_{17}=1$.
Example 18

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

Here, order does not matter. Therefore, we need to count combinations. There will be as many committees as there are combinations of 5 different persons taken 3 at a time. Hence, the required number of ways $={ }^5 \mathrm{C}_3=\frac{5!}{3!2!}=\frac{4 \times 5}{2}=10$.

Now, 1 man can be selected from 2 men in ${ }^2 \mathrm{C}_1$ ways and 2 women can be selected from 3 women in ${ }^3 \mathrm{C}_2$ ways. Therefore, the required number of committees

$
={ }^2 \mathrm{C}_1 \times{ }^3 \mathrm{C}_2=\frac{2!}{1!1!} \times \frac{3!}{2!1!}=6 .
$

Example 19

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 colour?

Solution

There will be as many ways of choosing 4 cards from 52 cards as there are combinations of 52 different things, taken 4 at a time. Therefore
$
\begin{aligned}
\text { The required number of ways }={ }^{52} \mathrm{C}_4=\frac{52!}{4!48!} & =\frac{49 \times 50 \times 51 \times 52}{2 \times 3 \times 4} \\
& =270725
\end{aligned}
$
(i) There are four suits: diamond, club, spade, heart and there are 13 cards of each suit. Therefore, there are ${ }^{13} \mathrm{C}_4$ ways of choosing 4 diamonds. Similarly, there are ${ }^{13} \mathrm{C}_4$ ways of choosing 4 clubs, ${ }^{13} \mathrm{C}_4$ ways of choosing 4 spades and ${ }^{13} \mathrm{C}_4$ ways of choosing 4 hearts. Therefore
$
\begin{aligned}
\text { The required number of ways } & ={ }^{13} \mathrm{C}_4+{ }^{13} \mathrm{C}_4+{ }^{13} \mathrm{C}_4+{ }^{13} \mathrm{C}_4 . \\
& =4 \times \frac{13!}{4!9!}=2860
\end{aligned}
$

(ii) There are 13 cards in each suit.
Therefore, there are ${ }^{13} \mathrm{C}_1$ ways of choosing 1 card from 13 cards of diamond, ${ }^{13} \mathrm{C}_1$ ways of choosing 1 card from 13 cards of hearts, ${ }^{13} \mathrm{C}_1$ ways of choosing 1 card from 13 cards of clubs, ${ }^{13} \mathrm{C}_1$ ways of choosing 1 card from 13 cards of spades. Hence, by multiplication principle, the required number of ways
$
={ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1=13^4
$
(iii) There are 12 face cards and 4 are to be selected out of these 12 cards. This can be done in ${ }^{12} \mathrm{C}_4$ ways. Therefore, the required number of ways $=\frac{12!}{4!8!}=495$.

(iv) There are 26 red cards and 26 black cards. Therefore, the required number of
$
\text { ways } \begin{aligned}
={ }^{26} \mathrm{C}_2 & \times{ }^{26} \mathrm{C}_2 \\
& =\left(\frac{26!}{2!24!}\right)^2=(325)^2=105625
\end{aligned}
$
(v) 4 red cards can be selected out of 26 red cards in ${ }^{26} \mathrm{C}_4$ ways. 4 black cards can be selected out of 26 black cards in ${ }^{26} \mathrm{C}_4$ ways.

Therefore, the required number of ways $={ }^{26} \mathrm{C}_4+{ }^{26} \mathrm{C}_4$
$
=2 \times \frac{26!}{4!22!}=29900 \text {. }
$