NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

Class 11 Permutation and Combinations Exercise 6.4 Solutions

Q1. ⁿC₈ = ⁿC₂, find ⁿC2.

A.1. ⁿC₈ = ⁿC₂

As, ⁿC_{a =} ⁿC_b

=>a = b or a = n – b

=>n = a + b

We have,

ⁿC₈ = ⁿC₂

=>n = 8 + 2

=>n = 10

Therefore,

ⁿC₂

= ⁿC₂

=  $\frac{10!}{2!\left(10-2\right)!}$

=  $\frac{10!}{2!8!}$

=  $\frac{1\mathrm{0\prime }<br>\mathrm{9\prime }<br>8!}{2!8!}$

=  $\frac{1\mathrm{0\prime }<br>9}{2}$

= 45

Q2. Determine n if

²ⁿC₃ : ⁿC₃ = 12 : 1(ii)²ⁿC₃ : ⁿC₃ = 11 : 1

A.2. i.²ⁿC₃ : ⁿC₃ = 12 : 1

=>  $\frac{2n!}{3!\left(2n-3\right)!}$ ÷  $\frac{n!}{3!\left(n-3\right)!}$ =  $\frac{12}{1}$

=>  $\frac{2n\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)!}{3!\left(2n-3\right)!}$ ×  $\frac{3!\left(n-3\right)!}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}$ = 12

=>  $\frac{2n\left(2n-1\right)\left(2n-2\right)}{n\left(n-1\right)\left(n-2\right)}$ = 12

=>  $\frac{2\left(2n-1\right)2\left(n-1\right)}{\left(n-1\right)\left(n-2\right)}$ = 12

=> 4(2n - 1) = 12(n – 2)

=> 8n – 4 = 12n – 24

=> 24 – 4 = 12n – 8n

=> 20 = 4n

=>n =  $\frac{20}{4}$

=>n = 5

ii.²ⁿC₃ : ⁿC₃ = 11 : 1

=>  $\frac{2n!}{3!\left(2n-3\right)!}$ ÷  $\frac{n!}{3!\left(n-3\right)!}$ =  $\frac{11}{1}$

=>  $\frac{2n\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)!}{3!\left(2n-3\right)!}$ ×  $\frac{3!\left(n-3\right)!}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}$ = 11

=>  $\frac{2n\left(2n-1\right)2\left(n-1\right)}{n\left(n-1\right)\left(n-2\right)}$ = 11

=> 4(2n – 1) = 11(n – 2)

=> 8n – 4 = 11n – 22

=> 22 – 4 = 11n – 8n

=> 18 = 3n

=>n =  $\frac{18}{3}$

=>n = 6

A.3. A chord is drawn by connecting 2 points on a circle.

As we are given with 21 points on the circle, we have the following combination to find the number of chords.

²¹C₂ = $\frac{21!}{2!\left(21-2\right)!}$ =  $\frac{2\mathrm{1\prime }<br>2\mathrm{0\prime }<br>19!}{\mathrm{2\prime }<br>\mathrm{1\prime }<br>19!}$ = 210

A.4.The number of ways of selecting a team consisting of 3 boys from 5 boys and 3 girls from 4 girls is

⁵C₃×⁴C₃

=  $\frac{5!}{3!\left(5-3\right)!}$ ×  $\frac{4!}{3!\left(4-3\right)!}$

=  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!\prime <br>2!}$ ×  $\frac{\mathrm{4\prime }<br>3!}{3!\prime <br>1!}$

=  $\frac{20}{2}$ ×  $\frac{4}{1}$

= 40

A.5. Since we are to select 3 balls from each colour in order to select 9 balls from the collections of 6, 5 and 5 balls of red, white and blue colours respectively, we can have the combination

⁶C₃(red) ×⁵C₃ (white) ×⁵C₃ (blue)

=  $\frac{6!}{3!\left(6-3\right)!}$ ×  $\frac{5!}{3!\left(5-3\right)!}$ ×  $\frac{5!}{3!\left(5-3\right)!}$

=  $\frac{\mathrm{6\prime }<br>\mathrm{5\prime }4\prime <br>3!}{\mathrm{3\prime }<br>\mathrm{2\prime }<br>3!}$ ×  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!<br>2!}$ ×  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!<br>2!}$

= 20 × 10 × 10

= 2000

A.6. In a deck of 52 cards there are 4 ace cards. The required number of ways of selecting one ace card from the four = ⁴C₁ = $\frac{4!}{1!\left(4-1\right)!}$ =  $\frac{4!}{3!}$ =  $\frac{4\times 3!}{3!}$ = 4

After selecting one ace we need to select the remaining 4 card from the remaining 48 card to have a combination of 5 cards. The required number of ways

= ⁴⁸C₄

=  $\frac{48!}{4!\left(48-4\right)!}$

=  $\frac{4\mathrm{8\prime }<br>4\mathrm{7\prime }<br>4\mathrm{6\prime }<br>4\mathrm{5\prime }<br>44!}{4!\prime <br>44!}$

=  $\frac{4\mathrm{8\prime }<br>4\mathrm{7\prime }<br>4\mathrm{6\prime }<br>45}{\mathrm{4\prime }<br>\mathrm{3\prime }<br>2}$

= 1,94,580

Therefore, the total number of ways for selecting 5 card combination out of a deck of 52 cards if there is exactly one ace in each combination

= ⁴C₁×⁴⁸C₄

= 4 × 1,94,580

= 7,78,320

A.7. We are given 17 players of which 5 players can bowl and 17 – 5 = 12 can bat. But we need to select a team of 11 in which there are exactly 4 bowlers.

Hence, the required number of ways

=⁵C₄ (bowl) x ¹²C_(11-4) (bat)

=  $\frac{5!}{4!\left(5-4\right)!}$ ×  $\frac{12!}{7!\left(12-7\right)!}$

=  $\frac{\mathrm{5\prime }<br>4!}{4!\prime <br>1!}$ ×  $\frac{1\mathrm{2\prime }<br>1\mathrm{1\prime }<br>1\mathrm{0\prime }<br>\mathrm{9\prime }<br>\mathrm{8\prime }<br>7!}{7!\prime <br>5!}$

= 5 ×  $\frac{1\mathrm{2\prime }<br>1\mathrm{1\prime }<br>1\mathrm{0\prime }<br>\mathrm{9\prime }<br>8}{\mathrm{5\prime }<br>\mathrm{4\prime }<br>\mathrm{3\prime }<br>\mathrm{2\prime }<br>1}$

= 5 × 11 × 9 × 8

= 3960

A.8. We are given 5 black and 6 red balls of which 2 black and 3 red balls can be selected.

Thus the required number of ways

= ^5C₂×⁶C₃

=  $\frac{5!}{2!\left(5-2\right)!}$ ×  $\frac{6!}{3!\left(6-3\right)!}$

=  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{\mathrm{2\prime }<br>3!}$ ×  $\frac{\mathrm{6\prime }<br>\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!\prime <br>\mathrm{3\prime }<br>2}$

= 10 × 20

= 200

A.9. A student can choose 5 out of 9 available courses. However 2 specific courses are made compulsory.

So, now a student has 3 choices out of the remaining 7 courses.

Therefore, the required number of ways

=⁷C₃

=  $\frac{7!}{3!\left(7-3\right)!}$

=  $\frac{\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>4!}{\mathrm{3\prime }<br>\mathrm{2\prime }<br>4!}$

= 35

Permutation and Combinations Miscellaneous Exercise Solutions

Q1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER ?

A.1. In the word DAUGHTER, there are 3 vowels, namely A, U, E and 5 consonants, namely D, G, H, T, R.

The number of ways of selecting 2 vowels out of 3

= ³C₂

=  $\frac{3!}{2!\left(3-2\right)!}$

=  $\frac{\mathrm{3\prime }<br>2!}{2!\prime <br>1!}$

= 3

The number of ways of selecting 3 consonants out of 5

= ⁵C₃

=  $\frac{5!}{3!\left(5-3\right)!}$

=  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!\prime <br>2!}$

= 5 × 2

= 10

Therefore, the number of combination of 3 consonants and 2 vowels is 3 × 10 = 30.

Each of these 30 combinations has 5 letters which can be arranged among themselves in 5! Ways.

Therefore, the required numbers of different words is

= 30 × 5! = 30 × 5 × 4 × 3 × 2 × 1 = 3600

A.2. In the word EQUATION, there are vowels (E, U, A, I, O) and 3 consonants (Q, T, N).

Treating vowels as a whole as 1st object and consonants as a whole as 2nd object, we can have an arrangement of 2! = 2.

Similarly, arrangement within the vowels = 5! = 5 × 4 × 3 × 2 × 1 = 120

And arrangement within the consonants = 3! = 3 × 2 × 1 = 6

Therefore, total number of possible arrangement = 2 × 120 × 6 = 1440

A.3. In the eleven-letter word EXAMINATION there are 2A’s, 2I’s and 2N’s and the rest are all different.

Since in a dictionary the words are listed according to the English alphabet we can only find words starting with A(as B, C, D are not a part of the letters forming the word EXAMINATION) listed before E.

Hence after fixing one A as first word we can rearrange the remaining 10 letters of which 2 are I, 2 are N and rest are all different.

Therefore, the required number of ways =  $\frac{10!}{2! 2!}$

=  $\frac{1\mathrm{0\prime }<br>\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>\mathrm{4\prime }<br>\mathrm{3\prime }<br>2!}{\mathrm{2\prime } <br>2!}$

= 10 × 9 × 8 × 7 × 6 × 5 × 2 × 3

= 9,07,200

A.4. Since the 6-digit numbers to be formed from the digits 0, 1, 3, 5, 7 and 9 has to be divisible by 10 we have to fix the unit place as 0. Now, the remaining 5 places can be filled only by the digits 1, 3, 5, 7 and 9.

Therefore, the required number of ways

= 5!

= 5 × 4 × 3 × 2 × 1

= 120

A.5. In an English word there are 5 vowels and 21 consonants.

The number of ways of selecting 2 vowel out of 5 = ⁵C₂

=  $\frac{5!}{2!\left(5-2\right)!}$

=  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{\mathrm{2\prime } <br>\mathrm{1\prime }<br>3!}$

= 5 × 2 = 10

The number of ways of selecting 2 consonants out of 21 = ²¹C₂

=  $\frac{21!}{2!\left(21-2\right)!}$

=  $\frac{2\mathrm{1\prime } <br>2\mathrm{0\prime }<br>19!}{\mathrm{2\prime } <br>\mathrm{1\prime }<br>19!}$

= 21 × 10

= 210

Therefore, the number of combinations of 2 vowels and 2 consonants is 10 × 210 = 2100

Each of these 2100 combinations has 4 letters which can be rearranged among themselves in 4! Ways.

Therefore, the required number of ways

= 4! × 2100

= 4 × 3 × 2 × 1 × 2100

= 50400

A.6. Since out of 8 total questions at least 3 questions has to be attempted from each of part I and II containing 5 and 7 questions respectively we can have the choices.

(a) 3 questions from I and 5 questions from II selected in ⁵C₃×⁷C₅ ways.

(b) 4 questions from I and 4 questions from II selected in ⁵C₄×⁷C₄ ways.

(c) 5 questions from I and 3 questions from II selected in ⁵C₅×⁷C₃ ways.

Therefore, the required number of ways.

= (⁵C₃×⁷C₅) + (⁵C₄×⁷C₄) + (⁵C₅×⁷C₃)

=  $\frac{5!}{3!\left(5-3\right)!}$ ×  $\frac{7!}{5!\left(7-5\right)!}$ +  $\frac{5!}{4!\left(5-4\right)!}$ ×  $\frac{7!}{4!\left(7-4\right)!}$ +  $\frac{5!}{5!\left(5-5\right)!}$ ×  $\frac{7!}{3!\left(7-3\right)!}$

=  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!\prime <br>2!}$ ×  $\frac{\mathrm{7\prime } <br>\mathrm{6\prime }<br>5!}{5!\prime <br>2!}$ +  $\frac{\mathrm{5\prime }<br>4!}{4!\prime <br>1!}$ ×  $\frac{\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>4!}{4!\prime <br>3!}$ +  $\frac{5!}{5!\prime <br>0!}$ ×  $\frac{\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>4!}{\mathrm{3\prime }<br>\mathrm{2\prime }<br>\mathrm{1\prime }<br>4!}$

=  $\frac{\mathrm{5\prime }<br>4}{\mathrm{2\prime } <br>1}$ ×  $\frac{\mathrm{7\prime } <br>6}{\mathrm{2\prime }<br>1}$ +  $\frac{5}{1}$ ×  $\frac{\mathrm{7\prime }<br>\mathrm{6\prime }<br>5}{\mathrm{3\prime }<br>\mathrm{2\prime }<br>1}$ + 1 × 35

= (10 × 21) + (5 × 35) + 35

= 210 + 175 + 35

= 420

A.7. In a deck of 52 card there are four kings.

So, number of ways of selecting exactly one king is ⁴C_1.

Now, after fixing one king card, we need to have the remaining 4 out of 5 cards to be a non-king i.e., only from the other 48 cards. So, number of ways of selecting is ⁴⁸C₄

Therefore, the required number of ways

= ⁴C₁×⁴⁸C₄

A.8. As out of the total 9 seats 4 women are to be at even places we can have the following arrangement.

Seat places

Also from this arrangement the women and men can rearrange among themselves.

Therefore, the required number of ways = 4! × 5!

= (4 × 3 × 2 × 1) × (5 × 4 × 3 × 2 × 1)

= 24 × 120

= 2880

A.9. In a class of 25 students, 10 students are to be selected for excursion. As 3 students decided that either all of them will join or none of them will join we have the options:

For the 3 students to be selected along with 7 other students from the remaining 25 – 3 = 22 students. This can be done in ³C₃ײ²C₇ ways.

For the 3 students to not be selected so that all 10 students will be from the remaining 25 – 3 = 22 students. This can be done in ³C₀ײ²C₁₀ ways.

Therefore, the required number of ways

= ³C₃× ²²C₇ + ³C₀ײ²C₁₀

= ²²C₇ + ²²C₁₀

A.10. In the 13 letter word ASSASSINATION there are 3-A, 4-S, 2-I, 2-N, 1-T and 1-O.

Since all the S are to be occurred together we treat them i.e. (SSSS) as single object. This single object together with 13 – 4 = 9 remaining object will account for 10 objects having 3-A, 2-I, 2-N, 1-T and 1-O and can be rearranged in  $\frac{10!}{3!2!2!}$

=  $\frac{1\mathrm{0\prime }<br>\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!\prime <br>\mathrm{2\prime }<br>1\mathrm{2\prime }<br>1}$

= 10 × 9 × 8 × 7 × 6 × 5

= 151200