CBSE Class 11 Maths Revision Notes Chapter 7

Q.1 In how many ways 7 pictures can be hung from 5 nails on a wall?

Ans

$\begin{array}{l}\text{Required number of ways}=\mathrm{P}_5^7\\ \mathrm{P}_5^7=\frac{7!}{(7-5)!}=\frac{7\times 6\times 5\times 4\times 3\times 2!}{2!}=7\times 6\times 5\times 4\times 3=2520\end{array}$

Q.2 How many different signals can be made using 3 flags at a time from 5 flags of different colours?

Ans

$\begin{array}{l}\text{Required number of signals}\\ \mathrm{P}_3^5=\frac{5!}{(5-3)!}=\frac{5\times 4\times 3\times 2!}{2!}=5\times 4\times 3=60\end{array}$

Q.3 Evaluate ⁵P₅ .

Ans

$\mathrm{p}_5^5=\frac{5!}{(5–5)!}=\frac{5\times 4\times 3\times 2\times 1}{0!}=\frac{120}{1}=120$

Q.4 Find the LCM of 3!, 4! and 5!.

Ans

5! = 5 × 4 × 3!
4! = 4 × 3!
3! = 3!
∴ LCM = 5 × 4 × 3! = 120

Q.5 A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of

(i) Exactly 3 girls?
(ii) at least 3 girls?
(iii) at most 3 girls?

Ans

$\begin{array}{l}\text{(i) A committee cnsisting of}3\text{girls and}4\text{boys can be formed in}\\ {}^4{\mathrm{C}}_3\times {}^9{\mathrm{c}}_4\text{ways}={}^4{\mathrm{c}}_1\times {}^9{\mathrm{c}}_4\\ =\frac{4}{1}\times \frac{9\times 8\times 7\times 6}{1.2.3.4}\text{ways}\\ =504\text{ways}\\ \text{(ii) A committee having at least}3\text{girls will consists of}\\ \text{(a)\hspace{0.33em}3 girls 4 boys}\\ \text{(b)\hspace{0.33em}4 girls 3 boys}\\ {\text{This can be done in}}^4{\mathrm{C}}_3\times {}^9{\mathrm{C}}_4+{}^4{\mathrm{C}}_4\times {}^9{\mathrm{C}}_3\text{ways}\\ =\frac{4}{1}\times \frac{9\times 8\times 7\times 6}{1\times 2\times 3\times 4}+1\times \frac{9\times 8\times 7}{1\times 2\times 3}\text{ways}\\ =504+84\text{ways}\\ =588\text{ways}\\ \text{(iii) A committee having at most}3\text{girls will consist of}\\ \text{(i) No girl}7\text{boys}\\ \text{(ii)}1\text{girl,}6\text{boys}\\ \text{(iii)}2\mathrm{girl},5\text{boys}\\ \text{(iv)}3\text{girl,}4\text{boys}\\ ={}^9{\mathrm{C}}_7+{}^4{\mathrm{C}}_1\times {}^9{\mathrm{C}}_6+{}^4{\mathrm{C}}_2\times {}^9{\mathrm{C}}_5+{}^9{\mathrm{C}}_3\times {}^9{\mathrm{C}}_4\text{ways}\\ ={}^9{\mathrm{C}}_7+{}^9{\mathrm{C}}_2+{}^4{\mathrm{c}}_1\times {}^9{\mathrm{C}}_3+{}^4{\mathrm{C}}_2+{}^4{\mathrm{C}}_1\times {}^9{\mathrm{C}}_4\text{ways}\\ =\frac{9\times 8}{1\times 2}+\frac{4}{1}\times \frac{9\times 8\times 7}{1\times 2\times 3}+\frac{4\times 3}{1\times 2}\times \frac{9\times 8\times 7\times 6}{1\times 2\times 3}+\frac{4}{1}\times \frac{9\times 8\times 7\times 6}{1\times 2\times 3\times 4}\\ =36+4\times 84+6\times 126+4\times 126\text{ways}\\ =36+336+1260\text{ways}\\ \text{=\hspace{0.33em}1632}\end{array}$

Q.6 In how many ways can the letters of the word PERMUTATIONS be arranged if the

(i) word starts with P and ends with S?
(ii) vowels are together?
(iii) there are always 4 letters between P and S?

Ans

Q.7 From a class of 15 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

Ans

There are two cases.
(a) If the 3 students join the excursion party then the number of combinations will be C₁= C(12, 7)
(b) If the 3 students do not join the excursion party. Then the number of combinations C₂= C(12, 10)
If C is the combination of choosing the excursion party, then

$\begin{array}{l}{\text{C\hspace{0.33em}=\hspace{0.33em}C}}_1{\text{\hspace{0.33em}+\hspace{0.33em}C}}_2\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}C\hspace{0.33em}(12,\hspace{0.33em}7)\hspace{0.33em}+\hspace{0.33em}C\hspace{0.33em}(12,\hspace{0.33em}10)}\\ =\frac{12!}{7!5!}+\frac{12!}{10!2!}\\ =\frac{12\times 11\times 10\times 9\times 8}{5\times 4\times 3\times 2\times 1}+\frac{12\times 11}{2}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}858}\end{array}$

Q.8 A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has

(i) no girl?
(ii) at least one boy and one girl?
(iii) at least 3 girls?

Ans

(i) Since the team will not include any girl therefore only boys are to be selected. 5 boys out of 7 boys can be selected in ⁷C₅ ways.
Therefore, the required number of ways

${=}^7{\mathrm{C}}_5=\frac{7!}{5!2!}=\frac{6\times 7}{2}=21$

(ii) Since at least one boy and one girl are to be there in every team. Therefore the team can consist of
(a) 1 boy and 4 girls
(b) 2 boys and 3 girls
(c) 3 boys and 2 girls
(d) 4 boys and 1 girls

1 boy and 4 girls can be selected in ⁷C₁ × ⁴C₄ ways
2 boys and 3 girls can be selected in ⁷C₂ × ⁴C₃ ways
3 boys and 2 girls can be selected in ⁷C₃ × ⁴C₂ ways
4 boys and 1 girl can be selected in ⁷C₄ × ⁴C₁ ways
Therefore, the required number of ways
= ⁷C₁ × ⁴C₄ + ⁷C₂ × ⁴C₃ + ⁷C₃ × ⁴C₂ + ⁷C₄ × ⁴C₁
= 7 + 84 + 210 + 140
= 441

(iii) Since the team has to consist of at least 3 girls. Therefore, the team can consist of
(a) 3 girls and 2 boys or
(b) 4 girls and 1 boy
Note that the team cannot have all 5 girls because, the group has only 4 girls.
3 girls and 2 boys can be selected in ⁴C₃ × ⁷C₂ ways
4 girls and 1 boy can be selected in ⁴C₄ × ⁷C₁ ways
Therefore the required number of ways = ⁴C₃ × ⁷C₂ + ⁴C₄ × ⁷C₁ = 84 + 7 = 91

Q.9 What is the number of ways of choosing 2 cards from a pack of 52 playing cards? In how many of these

(i) two cards are of the same suit?
(ii) two cards belong to different suits?
(iii) are face cards?
(iv) one is red card and one is black card?
(v) cards are of the same colour?

Ans

Q.10 In an examination a question paper consists of 12 questions divided into two parts, i.e., Part 1 and Part II, containing 5 and 7 questions respectively. A student is required to attempt 8 question in all, selecting at least 3 form each part. In how many ways can a student select the questions?

Ans

Student may select 8 question according to following-scheme
I (5 questions) II (7 question)
(a) 3 5
(b) 4 4
(c) 5 3
If P is the required number of ways, then

$\begin{array}{l}\mathrm{p}=\mathrm{C}(5,3)\times \mathrm{C}(7,5)+\mathrm{C}(5,4)\times \mathrm{C}(7,4)+\mathrm{C}(5,5)\times \mathrm{C}(7,3)\\ =\frac{5!}{3!2!}\times \frac{7!}{5!3!}+\frac{5!}{4!1!}\times \frac{7!}{4!3!}+\frac{5!}{5!0!}\times \frac{7!}{3!4!}\\ =\frac{5\times 4\times 3!}{2\times 1\times 3!}\times \frac{7\times 6\times 5!}{3\times 2\times 1\times 5!}+\frac{5\times 4!}{4!}\times \frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}+\frac{5\times 7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}\\ =10\times 7+5\times 35+5\times 35\\ =70+175+175=420\text{ways}\end{array}$

Q.11 In how many of the distinct permutations of the letters in MISSISSIPPI do the four ‘I’s not come together?

Ans

Q.12 How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4?

Ans

$\begin{array}{l}\text{The numbers for}7\text{-digit arrangements, clearly are,\hspace{0.33em}}\frac{7!}{3!2!}=420\\ \text{But, this will include those numbers also, which have}0\text{at at the extreme left position.}\\ \text{The number of such arrangements}\frac{6!}{3!2!}\text{(by fixing}0\text{at the extreme left position})=60\\ \text{Therefore, the required number of numbers}=420-60=360\end{array}$

Q.13 In how many ways can 6 girls and 4 boys be seated in a row so that no two boys are together?

Ans

Let us first seat the girls. This can be done in 6! ways. For each such arrangement, the four boys can be seated only at the cross marked places.
× G × G × G × G × G × G ×
There are 7 cross marked places and the four boys can be seated in ⁷P₄ ways. Hence, by multiplication principle, the total numbers of ways are
6! × ⁷P₄, i.e., 604800.

Q.14

${\mathbf{\text{Show that:}}}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}}\mathbf{+}{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}\mathbf{-}\mathbf{1}}\mathbf{=}{}^{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{C}}_{\mathbf{r}}$

Ans

$\begin{array}{l}\text{We\hspace{0.33em}have,}\\ {}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}+{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}+\frac{\mathrm{n}!}{(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r}+1)!}\\ =\frac{\mathrm{n}!}{\mathrm{r}\times (\mathrm{r}-1)!(\mathrm{n}-\mathrm{r})!}+\frac{\mathrm{n}!}{(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r}+1)(\mathrm{n}-\mathrm{r})!}\\ =\frac{\mathrm{n}!}{(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r})!}\left[\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{n}-\mathrm{r}+1}\right]\\ =\frac{\mathrm{n}!}{(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r})!}\times \frac{\mathrm{n}-\mathrm{r}+1+\mathrm{r}}{\mathrm{r}(\mathrm{n}-\mathrm{r}+1)}=\frac{(\mathrm{n}+1)!}{\mathrm{r}!(\mathrm{n}+1-\mathrm{r})!}\\ ={}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}}\end{array}$

Q.15 How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 5, 6, if no digit is repeated?

Ans

Q.16

$\mathbf{\text{Show that}}\mathbf{r}\mathbf{\text{does not have integral value for}}\mathbf{6}{}^{\mathbf{5}}{\mathbf{P}}_{\mathbf{r}}\mathbf{=}\mathbf{7}{}^{\mathbf{6}}{\mathbf{P}}_{\mathbf{r}\mathbf{-}\mathbf{1}}\mathbf{.}$

Ans

$\begin{array}{l}\text{We have}6{}^5{\mathrm{P}}_{\mathrm{r}}=7{}^6{\mathrm{P}}_{\mathrm{r}-1}\\ \text{or\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}6\times \frac{5!}{(5-\mathrm{r})!}=7\times \frac{6!}{(6-\mathrm{r}+1)!}\\ \mathrm{or}\frac{6!}{(5-\mathrm{r})!}=\frac{7\times 6!}{(6-\mathrm{r}+1)(6-\mathrm{r})(6-\mathrm{r}-1)!}\\ \text{or}(7-\mathrm{r})(6-\mathrm{r})=7\\ \text{or \hspace{0.33em}}{\mathrm{r}}^2-13\mathrm{r}+35=0,\text{which does not give integral value of}\mathrm{r}\text{.}\end{array}$

Q.17

$\mathbf{\text{Find}}\mathbf{n}\mathbf{}\mathbf{:}{}^{\mathbf{n}}{\mathbf{p}}_{\mathbf{5}}\mathbf{=}\mathbf{12}\mathbf{P}_{\mathbf{3}\mathbf{!}}^{\mathbf{n}}\mathbf{n}\mathbf{>}\mathbf{4}$

Ans

$\begin{array}{l}{}^{\mathrm{n}}{\mathrm{p}}_5=12\mathrm{P}_{3!}^{\mathrm{n}}\mathrm{n}>4\\ \Rightarrow \frac{\mathrm{n}!}{(\mathrm{n}-5)!}=12\frac{\mathrm{n}!}{(\mathrm{n}-3)!}\\ \Rightarrow \frac{\bcancel{\mathrm{n}!}}{\left(\bcancel{\mathrm{n}-5}\right)!}=12\frac{\bcancel{\mathrm{n}!}}{(\mathrm{n}-3)(\mathrm{n}-4)\left(\bcancel{\mathrm{n}-5}\right)!}\\ \Rightarrow (\mathrm{n}-3)(\mathrm{n}-4)=12\\ \Rightarrow {\mathrm{n}}^2-3\mathrm{n}-4\mathrm{n}+12-12=0\\ \Rightarrow {\mathrm{n}}^2-7\mathrm{n}=0\\ \Rightarrow \mathrm{n}=0\text{or}\mathrm{n}=7\\ \text{Hence}\mathrm{n}=7\text{\hspace{0.33em}(since}\mathrm{n}>4\text{)}\end{array}$

Q.18

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

Ans

$\begin{array}{l}\text{Here}\frac{1}{8!}+\frac{1}{9\times 8!}=\frac{\mathrm{x}}{10\times 9\times 8!}\\ \Rightarrow 1+\frac{1}{9}=\frac{\mathrm{x}}{90}\\ \Rightarrow \frac{10}{9}=\frac{\mathrm{x}}{90}\\ \Rightarrow \mathrm{x}=100\end{array}$

Q.19 How many 3-digit numbers can be formed from the digits 2, 3, 4 , 5 and 6 assuming that

(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?

Ans

(i) Total number of digits = 5
∴ Total 3-digit numbers when digits repetition allowed = 5 × 5 × 5 = 125

(ii) Total 3-digit numbers when digits repetition is not allowed is

$\mathrm{P}_3^5=\frac{5!}{(5-3)!}=5\times 4\times 3=60$

Q.20

$\mathbf{\text{Compute}}\frac{\mathbf{10}\mathbf{!}}{\mathbf{7}\mathbf{!}\mathbf{}\mathbf{3}\mathbf{!}}\mathbf{\text{.}}$

Ans

$\frac{10!}{7!3!}=\frac{10\times 9\times 8\times \cancel{7!}}{\cancel{7!}\times 3\times 2\times 1}=\frac{10\times 9\times 8}{3\times 2}=120$

Q.21 How many permutations of the letters of the word ‘APPLE’ are possible?

Ans

Total letters = 5
A = 1 letter
P = 2 letters
L = 1 letter
E = 1 letter

$\text{Permutation}=\frac{5!}{1!\times 2!\times 1!\times 1!}=\frac{5\times 4\times 3\times 2!}{2!}=5\times 4\times 3=60$

Q.22 Find the number of diagonals in polygon of 14 sides.

Ans

$\begin{array}{l}\text{Since a polygon has}14\text{sides, therefore, there are\hspace{0.33em}}14\text{non collinear points.}\\ \text{For drawing diagonal, we can select any}2\text{points}\\ \text{out of}14{\text{in}}^{14}{\mathrm{C}}_2\text{ways}=\frac{14!}{2!(14-2)!}\\ =\frac{14!}{2!12!}\\ =\frac{14\times 13\times 12!}{2\times 12!}\\ =\frac{14\times 13}{2}\\ =7\times 13\\ =91\text{ways}\\ \text{But, this selection of points includes points of sides}\left(14\right)\text{also.}\\ \text{Therefore, the number of diagonals}=91-14\\ =77\end{array}$

Q.23 If ⁿC₇ = ⁿC₄, find the value of n.

Ans

We know that ⁿC_x = ⁿC_y ⇒ n = x + y or x = y.
So, for given ⁿC₇ = ⁿC₄ , we get
n = 7 + 4
= 11

Q.24 Evaluate ¹⁰⁰C₉₈.

Ans

¹⁰⁰C₉₈ = (100)(99)/2
= (50)(99)
= 4950

Q.25 If ⁿP_r = 720 and ⁿC_r = 120, find r.

Ans

$\begin{array}{l}\text{We know that}\\ {}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{{}^{\mathrm{n}}{\mathrm{P}}_{\mathrm{r}}}{\mathrm{r}!}\\ \Rightarrow 120=\frac{720}{\mathrm{r}!}\\ \Rightarrow \mathrm{r}!=6\\ \Rightarrow \mathrm{r}!=3!\\ \Rightarrow \mathrm{r}=3\end{array}$

Q.26 From a group of 15 cricket players, a team of 11 players is to be chosen. In how many ways can this be done?

Ans

$\begin{array}{l}\text{Required number of ways}={}^{15}{\mathrm{C}}_{11}\\ =\frac{15!}{4!\times 11!}\\ =\frac{12\times 13\times 14\times 15}{2\times 3\times 4}\\ =15\times 7\times 13\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=1365}\end{array}$

Q.27 A round table conference is to be held between the 7 delegates of two countries. Find the number of ways in which they can be seated if 2 particular delegates are never to sit together.

Ans

Since the number of circular permutation of n different things = (n – 1)!
Therefore, the number of ways in which 7 delegates can be seated on round table = (7 – 1)! = 6! ways
= 6 × 5 × 4 × 3 × 2 × 1
= 720
If 2 particular delegates like to seat together then the number of seating arrangement = (6 – 1)! × 2!
= 5! × 2
= 5 × 4 × 3 × 2 × 1 × 2
= 240
The number of ways in which they can be seated if 2 particular delegates are never to sit together = 720 – 240
= 480

Q.28 Find r if (i) ⁵P_r = ⁶P_r–1 (ii) ^n–1P_r = 2 ⁶P_r–1

Ans

$\begin{array}{l}{\text{(i)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}}^5{\mathrm{P}}_{\mathrm{r}}={}^6{\mathrm{P}}_{\mathrm{r}-1}\\ \frac{5!}{(5-\mathrm{r})!}=\frac{6!}{(6-\mathrm{r}+1)!}\\ \frac{5!}{(5-\mathrm{r})!}=\frac{6\times 5!}{(7-\mathrm{r})(6-\mathrm{r})(5-\mathrm{r})!}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(7}-\text{r)(6}-\text{r)\hspace{0.33em}=\hspace{0.33em}6}\\ 42-13\mathrm{r}+{\mathrm{r}}^2-6=0\\ {\mathrm{r}}^2-13\mathrm{r}+36=0\\ (\mathrm{r}-9)(\mathrm{r}-4)=0\\ \Rightarrow \mathrm{r}=9,4\\ \text{Neglecting}\mathrm{r}=9\\ \text{So, \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{r}=4.\\ {\text{(ii)\hspace{0.33em}\hspace{0.33em}}}^5{\mathrm{P}}_{\mathrm{r}}=2{}^6{\mathrm{P}}_{\mathrm{r}-1}\\ \frac{5!}{(5-\mathrm{r})!}=2\times \frac{6!}{(6-\mathrm{r}+1)!}\\ \frac{5!}{(5-\mathrm{r})!}=2\times \frac{6\times 5!}{(7-\mathrm{r})(6-\mathrm{r})(5-\mathrm{r})!}\\ (7-\mathrm{r})(6-\mathrm{r})=12\\ 42-13\mathrm{r}+{\mathrm{r}}^2=12\\ {\mathrm{r}}^2-13\mathrm{r}+42-12=0\\ {\mathrm{r}}^2-13\mathrm{r}+30=0\\ (\mathrm{r}-10)(\mathrm{r}-3)=0\\ \mathrm{r}=10,3\\ \mathrm{r}=3\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(Neglecting}\mathrm{r}=10)\end{array}$

Q.29 Find the number of different arrangements for letters of the word EQUATION. In how many ways, do the vowels come together?

Ans

There are 8 different letters in the word EQUATION
This can be arranged in 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 ways
= 40320 ways
There are 5 vowels and we will assume it as a single letter.
Now, bundle of 5 vowels and letters Q, T, N that is 4 different letters can be arranged in 4! = 4 x 3 x 2 x 1 = 24 ways.
The 5 different vowels can be arranged among itself in 5! = 5 × 4 × 3 × 2 × 1 = 120 ways.
Thus, the total number of arrangement in which vowels come together = 120 × 24 = 2880 ways.

Q.30 Find the value of n if (i) ²ⁿC₃ : ⁿC₃ = 12 : 1 (ii) ²ⁿC₃ : ⁿC₃ = 11:1

Ans

$\begin{array}{l}{\text{(i)}}^{2\mathrm{n}}{\mathrm{C}}_3:{}^{\mathrm{n}}{\mathrm{C}}_3=12:1\\ \frac{{}^{2\mathrm{n}}{\mathrm{C}}_3}{{}^{\mathrm{n}}{\mathrm{C}}_3}=12:1\Rightarrow \frac{\frac{2\mathrm{n}!}{3!(2\mathrm{n}-3)!}}{\frac{\mathrm{n}!}{3!(\mathrm{n}-3)!}}=\frac{12}{1}\\ \frac{2\mathrm{n}!}{(2\mathrm{n}-3)!}\times \frac{(\mathrm{n}-3)!}{\mathrm{n}!}=\frac{12}{1}\\ \frac{2\mathrm{n}(2\mathrm{n}-1)(2\mathrm{n}-2)(2\mathrm{n}-3)!}{(2\mathrm{n}-3)!}\times \frac{(\mathrm{n}-3)!}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)!}=\frac{12}{1}\\ \Rightarrow 2\mathrm{n}-1=3(\mathrm{n}-2)\\ \Rightarrow \mathrm{n}=5\\ {\text{(ii)}}^{2\mathrm{n}}{\mathrm{C}}_3:{}^{\mathrm{n}}{\mathrm{C}}_3=11:1\\ \frac{\mathrm{C}_3^{2\mathrm{n}}}{\mathrm{C}_3^{\mathrm{n}}}=\frac{11}{1}\Rightarrow \frac{\frac{2\mathrm{n}!}{3!(2\mathrm{n}-3)!}}{\frac{\mathrm{n}!}{3!(\mathrm{n}-3)!}}=\frac{11}{1}\\ \Rightarrow \frac{2\mathrm{n}!}{(2\mathrm{n}-3)!}\times \frac{(\mathrm{n}-3)!}{\mathrm{n}!}=\frac{11}{1}\\ \Rightarrow \frac{2\mathrm{n}(2\mathrm{n}-1)(2\mathrm{n}-2)(2\mathrm{n}-3)!}{(2\mathrm{n}-3)!}\times \frac{(\mathrm{n}-3)!}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)!}=\frac{11}{1}\\ \Rightarrow \frac{4(2\mathrm{n}-1)}{(\mathrm{n}-2)}=\frac{11}{1}\Rightarrow \mathrm{n}=6\end{array}$

Q.31 Every body in a room shakes hands with everybody else. The total number of handshakes is 66. Find the number of persons in the room.

Ans

$\begin{array}{l}\text{Let the number of persons in the room be}\mathrm{n}\text{\hspace{0.33em}Since one person at a time can make}\\ \text{hand shake with}1\text{person only.}\\ \text{The total number of hand shakes}=\text{The total\hspace{0.33em} number of selection of any}2\text{person}\\ \text{out of}\mathrm{n}\text{\hspace{0.33em}person}={}^{\mathrm{n}}{\mathrm{C}}_2\\ \because {}^{\mathrm{n}}{\mathrm{C}}_2=66\text{(given)}\\ \Rightarrow \frac{\mathrm{n}!}{2!(\mathrm{n}-2)!}=66\\ \Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)!}{2(\mathrm{n}-2)!}=66\\ \Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)}{2}=66\\ \Rightarrow \mathrm{n}(\mathrm{n}-1)=2\times 66\\ \Rightarrow \mathrm{n}(\mathrm{n}-1)=2\times 11\times 6\\ \Rightarrow \mathrm{n}(\mathrm{n}-1)=12\times 11\\ \text{On comparing factors, we get n\hspace{0.33em}=\hspace{0.33em}12}\\ \text{The number of persons in the room is}12.\end{array}$

Q.32 In CWG for a rugby championship, there were 153 matches played. Every two teams played one match with each other. Find the number of teams of rugby participating in the Commonwealth Games.

Ans

$\begin{array}{l}\text{Let the number of rugby team in CWG be}\mathrm{n}\\ \text{The total number of matches played = The total}\\ \text{number of selection of any}2\text{teams out of}\mathrm{n}\\ \text{teams}={}^{\mathrm{n}}{\mathrm{C}}_2\\ \because {}^{\mathrm{n}}{\mathrm{C}}_2=153\text{\hspace{0.33em}(given)}\\ \Rightarrow \frac{\mathrm{n}!}{2!(\mathrm{n}-2)!}=153\\ \Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)}{2}=153\\ \Rightarrow \mathrm{n}(\mathrm{n}-1)=2\times 153\\ \Rightarrow \mathrm{n}(\mathrm{n}-1)=2\times 17\times 9\\ \Rightarrow \mathrm{n}(\mathrm{n}-1)=18\times 17\\ \text{on comparing factors, we get}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}n\hspace{0.33em}=\hspace{0.33em}18}\\ \text{The number of rugby teams in the CWG is}18\end{array}$

Q.33 There are 10 professors and 20 students, out of whom a committee of 2 professors and 3 students is to be formed. Find the number of ways in which this can be done if a particular student is excluded.

Ans

$\begin{array}{l}\text{We have to make a committee of}5\text{persons.}\\ \text{First, we will select}2\text{professors out of}10\\ {\text{Professors in}}^{10}{\mathrm{C}}_2\text{ways}=\frac{10!}{2!(10-2)!}\\ =\frac{10!}{2!8!}\\ =\frac{10\times 9\times 8!}{2\times 8!}\\ =\frac{10\times 9}{2}\\ =5\times 9\\ =45\text{ways}\\ \text{and}3\text{students can be selected out of\hspace{0.33em}}(20-1)\text{since}1\text{is excluded}=19\text{students in}\\ {}^{19}{\mathrm{C}}_3\text{ways}=\frac{19!}{3!(19-3)!}\\ =\frac{19!}{3!16!}\\ =\frac{19\times 18\times 17\times 16!}{3\times 2\times 1\times 16!}\\ =19\times 3\times 17\\ =19\times 51\\ =969\text{ways}\\ \text{The number of ways in which}2\text{Professors and}3\text{students can be selected}=45\times 969\\ =43605\end{array}$

Q.34 Find the number of different arrangements for letters of the word EQUATION in which vowels do not come together.

Ans

There are 8 different letters in the word EQUATION.
This can be arranged in 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 ways
= 40320 ways
There are 5 vowels and we will assume it as a single letter.
Now, bundle of 5 vowels and letters Q,T,N that is 4 different letters can be arranged in 4! = 4 × 3 × 2 × 1 = 24 ways
The 5 different vowels can be arranged among itself in 5! = 5 × 4 × 3 × 2 × 1 = 120 ways
The total number of arrangement in which vowels come together = 120 × 24 = 2880 ways
Thus, the total number of arrangement in which vowels do not come together = Total no. of arrangements – Total no. of arrangements with vowels coming together
= 40320 ways – 2880 ways
= 37440 ways

Q.35 The value of 0! × 2! × 3! is ____.

Ans

0! × 2! × 3! = 1 × (2 × 1) × (3 × 2 × 1)
= 12
The value of 0! × 2! × 3! is 12.

Q.36

${\mathbf{\text{If}}}^{\mathbf{\text{n}}}{\mathbf{\text{C}}}_{\mathbf{\text{10}}}\mathbf{=}{}^{\mathbf{\text{n}}}{\mathbf{\text{C}}}_{\mathbf{\text{3}}}\mathbf{,}\mathbf{\text{then value of n is\_\_\_.}}$

Ans

$\begin{array}{l}{}^{\text{n}}{\text{C}}_{\text{10}}={}^{\text{n}}{\text{C}}_{\text{3}}\left[\mathrm{G}\text{iven}\right]\\ ={}^{\text{n}}{\text{C}}_{\text{n}-\text{3}}[\because {}^{\text{n}}{\text{C}}_{\text{r}}{=}^{\text{n}}{\text{C}}_{\text{n}-\text{r}}]\\ \Rightarrow 10=\mathrm{n}-3\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}n}=10+3\\ =13\\ {\text{If}}^{\text{n}}{\text{C}}_{\text{10}}={}^{\text{n}}{\text{C}}_{\text{3}},\text{then value of n is}\underset{¯}{\text{13}}\text{.}\end{array}$

Q.37 What are the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these, four cards belong to four different suits?

Ans

$\begin{array}{l}\text{There are number of ways of choosing}4\text{cards from}52\text{cards as there are combinations}\\ \text{of}52\text{different things, taken}4\text{at time.}\\ \text{Therefore, the required number of ways}\\ =\mathrm{C}_4^{52}=\frac{52!}{4!48!}=\frac{49\times 50\times 51\times 52}{2\times 3\times 4}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}270725}\\ \text{There are 13 cards in each suit.}\\ \text{Therefore, there are}\mathrm{C}_1^{13}\text{\hspace{0.33em}ways of choosing 1 card from 13 card of diamond,}\mathrm{C}_1^{13}\text{ways of}\\ \text{choosing 1 card from 13 cards of hearts,}\\ {}^{13}{\mathrm{C}}_1\text{ways of choosing}1\text{card from}13\text{cards of clubs.}\\ {}^{13}{\mathrm{C}}_1\text{ways of choosing}1\text{card from}13\text{cards of spades.}\mathrm{Hence},\mathrm{by}\text{\hspace{0.33em} multiplication principle,}\\ \text{the required number of ways}\\ ={}^{13}{\mathrm{C}}_1+{}^{13}{\mathrm{C}}_1+{}^{13}{\mathrm{C}}_1+{}^{13}{\mathrm{C}}_1={13}^4\end{array}$

Q.38 What are the number of ways of choosing 4 cards from a pack of 52 playing cards? Out of these, how many are face cards?

Ans

$\begin{array}{l}\text{There are number of ways of choosing}4\text{cards from}52\text{cards as there are combinations}\\ \text{of}52\text{different things, taken}4\text{at time.}\\ \text{Therefore, the required number of ways}\\ =\mathrm{C}_4^{52}=\frac{52!}{4!48!}=\frac{49\times 50\times 51\times 52}{2\times 3\times 4}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}270725}\\ \text{There are}12\text{face cards and}4\text{are to be selected out of these}12\text{cards.}\\ {\text{This can be done in}}^{12}{\mathrm{C}}_4\text{ways.}\\ \text{Therefore, the required number of ways}=\frac{12!}{4!8!}=485.\end{array}$

Q.39 How many different signals can be made using 3 flags at a time from 5 flags of different colours?

Ans

$\begin{array}{l}\text{Required number of signals}\\ {}^5{\mathrm{p}}_3=\frac{5!}{(5-3)!}=\frac{5\times 4\times 3\times 2!}{2!}=5\times 4\times 3=60\end{array}$

Q.40 If ⁿC₇ = ⁿC₄, find the value of n.

Ans

We know that ⁿC_x = ⁿC_y ⇒ n = x + y or x = y.
So, for given ⁿC₇ = ⁿC₄ , we get
n = 7 + 4
= 11

Q.41 Evaluate ¹⁰⁰C₉₈.

Ans

¹⁰⁰C₉₈ = (100)(99)/2
= (50)(99)
= 4950

Q.42 If ⁿP_r = 720 and ⁿC_r = 120, find r.

Ans

$\begin{array}{l}\text{We know that}\\ {}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{{\mathrm{np}}_{\mathrm{r}}}{\mathrm{r}!}\\ \Rightarrow 120=\frac{720}{\mathrm{r}!}\\ \Rightarrow \mathrm{r}!=6\\ \Rightarrow \mathrm{r}!=3!\\ \Rightarrow \mathrm{r}=3\end{array}$