NCERT Solutions Class 11 Maths Chapter 7

Q.1 How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that

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

Ans

The given three digits are 1, 2, 3, 4 and 5.
Number of digits in number to be formed = 3
(i) If repetition of digits is allowed.
Number of ways to fill one’s digit place = 5
Number of ways to fill ten’s digit place = 5
No. of ways to fill hundred’s digit place = 5
So, total formed 3-digit numbers = 5 × 5 × 5
= 125.
Thus, total 3-digit numbers are 125.

(ii) If repetition of digits is not allowed.
Number of ways to fill one’s digit place = 5
Number of ways to fill ten’s digit place = 4
No. of ways to fill hundred’s digit place = 3
So, total formed 3-digit numbers = 5 × 4 × 3
= 60
Thus, total 3-digit numbers are 60.

Q.2 How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Ans

The given three digits are 1, 2, 3, 4, 5 and 6.
Number of digits in number to be formed = 3
If repetition of digits is allowed.
Number of ways to fill one’s digit place = 3
Number of ways to fill ten’s digit place = 6
No. of ways to fill hundred’s digit place = 6
So, total formed 3-digit numbers = 6 x 6 x 3 = 108.
Thus, total 3-digit even numbers are 108.

Q.3 How many 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Ans

Number of letters of the English alphabet taken = 10
Number of letter used in making letter codes = 4
Since, no letter can be repeated in these codes.

So, number of ways to fill first place = 10
Number of ways to fill second place = 9
Number of ways to fill third place = 8
Number of ways to fill fourth place = 7
Total number of 4-letter codes = 10 x 9 x 8 x 7
= 5040.
Thus, total 4-letter codes are 5040.

Q.4 How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Ans

Total given digits from 0 to 9 = 10
Number of digits in telephone number = 5
Since, each number starts with 67. So, number of
vacant places in telephone number = 3

Number of ways to fill third place = 8
Number of ways to fill fourth place = 7
Number of ways to fill fifth place = 6
So, number of 5-digit telephone numbers
= 8 x 7 x 6
= 336
Thus, total number of 5-digit telephone numbers is 336.

Q.5 A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Ans

Number of possible outcomes in first toss of coin = 2
Number of possible outcomes in 2^nd toss of coin = 2
Number of possible outcomes in 3^rd toss of coin = 2
Total number of possible outcomes in 3 tosses of coin
= 2 × 2 × 2
= 8
Thus, total possible outcomes are 8.

Q.6 Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Ans

Number of flags of different colours = 5
Flags required for a signal = 2
Number of ways to take first flag for a signal
= 5
Number of ways to take second flag for a signal
= 4
Number of different signals generated by 5 flags
= 5 × 4
= 20
Thus, total number of signals generated by using 5 flags is 20.

Q.7 Evaluate: (i) 8! (ii) 4! – 3!

Ans

(i) 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320
(ii) 4! – 3! = 4 × 3 × 2 × 1 – 3 × 2 × 1
= 24 – 6
= 18

Q.8 Is 3! + 4! = 7! ?

Ans

3! + 4! = 3 × 2 × 1 + 4 × 3 × 2 × 1
= 6 + 24
= 30
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040
No, 3! + 4! ≠ 7!

Q.9

$\mathrm{Compute}:\frac{8!}{6!\times 2!}$

Ans

$\begin{array}{l}\frac{8!}{6!\times 2!}=\frac{\text{8}\times \text{7}\times \text{6}\times \text{5}\times \text{4}\times \text{3}\times \text{2}\times \text{1}}{\text{6}\times \text{5}\times \text{4}\times \text{3}\times \text{2}\times \text{1}\times \text{2}\times \text{1}}\\ =\frac{\text{8}\times \text{7}}{\text{2}\times \text{1}}\\ =28\end{array}$

Q.10

$\mathrm{If}\frac{1}{6!}+\frac{1}{7!}=\frac{\mathrm{x}}{8!},\mathrm{find}\mathrm{x}.$

Ans

$\begin{array}{l}\mathrm{We}\text{\hspace{0.33em}are\hspace{0.33em}given}\\ \frac{1}{6!}+\frac{1}{7!}=\frac{\text{x}}{8!}\\ \frac{1}{6!}+\frac{1}{7.6!}=\frac{\text{x}}{8!}\\ \frac{1}{6!}\left(1+\frac{1}{7}\right)=\frac{\text{x}}{8!}\\ \frac{1}{6!}\left(\frac{7+1}{7}\right)=\frac{\text{x}}{8!}\\ \frac{1}{6!}\left(\frac{8}{7}\right)=\frac{\text{x}}{8!}\\ \frac{8}{7!}=\frac{\text{x}}{8.\left(7!\right)}\\ 8=\frac{\text{x}}{8}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}x}=64\end{array}$

Q.11

$\begin{array}{l}\mathbf{Evaluate}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{–}\mathbf{r}\mathbf{)}\mathbf{!}}\mathbf{,}\mathbf{\text{}}\mathbf{when}\\ \mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{n}\mathbf{=}\mathbf{6}\mathbf{,}\mathbf{r}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{\left(}\mathbf{ii}\mathbf{\right)}\mathbf{}\mathbf{n}\mathbf{=}\mathbf{9}\mathbf{,}\mathbf{r}\mathbf{=}\mathbf{5}\mathbf{.}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{n}=6,\mathrm{r}=2\\ \frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{r})!}=\frac{6!}{(6-2)!}\\ =\frac{6\times 5\times 4!}{4!}\\ =30\\ \left(\mathrm{ii}\right)\mathrm{n}=9,\mathrm{r}=5\\ \frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{r})!}=\frac{9!}{(9-5)!}\\ \begin{array}{l}=\frac{9\times 8\times 7\times 6\times 5\times 4!}{4!}\\ =15120\end{array}\end{array}$

Q.12 How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Ans

Number of digits from 1 to 9 = 9
Number of digits in each number = 3
If no digit is repeated, then
Number of 3-digit numbers = ⁹P₃
= 9! /(9 – 3)!
= 9 × 8 × 7
= 504
Thus, there are 504, 3-digit numbers which can be formed by using the digits 1 to 9.

Q.13 How many 4-digit numbers are there with no digit repeated?

Ans

Number of digits from 0 to 9 = 10
Number of digits in each number = 4
If no digit is repeated, then
Number of 4-digit numbers = 9 × ⁹P₃
= 9 × 9 × 8 × 7
= 4536
Thus, there are 4536, 4-digit numbers which can be formed by using the digits 0 to 9.

Q.14 How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Ans

Number of given digits = 6
Number of digits in each number = 3
Since, each number is even, so last digit may be 2, 4 or 6.
If no digit is repeated, then Number of ways to fill remaining two places of 3-digit numbers = ⁵P₂
= 5! /(5 – 2)!
= 5! / 3!
= 20
Number of ways to arrange 3 even numbers at one’s place = 3
Total formed 3-digit even numbers = 20 × 3 = 60
Thus, there are 60, 3-digit even numbers.

Q.15 Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

Ans

Number of given digits = 5
Number of digits in each number = 4
Number of 4-digit numbers = ⁵P₄
= 5!
= 120
Now, we find 4 digit even numbers.
Only 2 numbers are possible at units place (2, 4) as we need even number.
If no digit is repeated, then number of ways to fill remaining three places of 4-digit numbers = ⁴P₃
= 4! /(4 – 3)!
= 4! / 1!
= 24
Number of ways to arrange 2 even numbers at one’s place = 2
Total formed 4-digit even numbers = 24 x 2
= 48
Thus, there are 48, 4 digit even numbers.

Q.16 From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?

Ans

$\begin{array}{l}\mathrm{Number}\text{of persons in committee}=8\\ \mathrm{Number}\text{of chairman to be selected}=\text{1}\\ \mathrm{Number}\text{of vice chairman to be selected}=\text{1}\\ \text{Number of persons to be selected for both posts}\\ =\text{2}\\ \text{Number of ways to select a chairman and vice chairman}\\ ={}^8{\mathrm{P}}_2\\ =\frac{8!}{(8-2)!}\\ =\frac{8\times 7\times 6!}{6!}\\ =56\\ \mathrm{Thus},\text{ the number of ways to choose both persons is 56.}\end{array}$

Q.17

$\mathbf{Find}\mathbf{}\mathbf{n}\mathbf{}\mathbf{if}{\mathbf{}}^{\mathbf{n}\mathbf{–}\mathbf{1}}{\mathbf{P}}_{\mathbf{3}}\mathbf{}\mathbf{:}{\mathbf{}}^{\mathbf{n}}{\mathbf{P}}_{\mathbf{4}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{:}\mathbf{}\mathbf{9}$

Ans

$\begin{array}{l}\mathrm{We}\text{\hspace{0.33em}are\hspace{0.33em}given:}\\ {}^{\text{n}-1}{\text{P}}_3:{}^{\text{n}}{\text{P}}_4=1:9\\ \Rightarrow \frac{{}^{\text{n}-1}{\text{P}}_3}{{}^{\text{n}}{\text{P}}_4}=\frac{1}{9}\\ \Rightarrow \frac{\left\{\frac{\left(\text{n}-1\right)!}{\left(\text{n}-1-3\right)!}\right\}}{\left\{\frac{\text{n}!}{\left(\text{n}-4\right)!}\right\}}=\frac{1}{9}\\ \Rightarrow \frac{\left(\text{n}-1\right)!}{\left(\text{n}-1-3\right)!}\times \frac{\left(\text{n}-4\right)!}{\text{n}!}=\frac{1}{9}\\ \Rightarrow \frac{\left(\text{n}-1\right)!}{\left(\text{n}-4\right)!}\times \frac{\left(\text{n}-4\right)!}{\text{n}\left(\text{n}-1\right)!}=\frac{1}{9}\\ \Rightarrow \frac{1}{\text{n}}=\frac{1}{9}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}n}=9\end{array}$

Q.18

$\begin{array}{l}\mathbf{\text{Find r if}}\\ \left(\text{i}\right){}^{\mathbf{\text{5}}}{\mathbf{\text{P}}}_{\mathbf{\text{r}}}{\mathbf{\text{=\hspace{0.33em}2\hspace{0.33em}}}}^{\mathbf{\text{6}}}{\mathbf{\text{P}}}_{\mathbf{\text{r}}\mathbf{–}\mathbf{\text{1}}}\\ \left(\text{ii}\right){}^{\mathbf{\text{5}}}{\mathbf{\text{P}}}_{\mathbf{\text{r}}}{\mathbf{\text{=\hspace{0.33em}}}}^{\mathbf{\text{6}}}{\mathbf{\text{P}}}_{\mathbf{\text{r–1}}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right){}^5{\mathrm{P}}_{\mathrm{r}}=2^6{\mathrm{P}}_{\mathrm{r}-1}\\ \Rightarrow \frac{5!}{(5-\mathrm{r})!}=2\cdot \frac{6!}{(6-\mathrm{r}+1)!}\\ \Rightarrow \frac{5!}{(5-\mathrm{r})!}=2\cdot \frac{6!}{(7-\mathrm{r})!}\\ \Rightarrow \frac{5!}{(5-\mathrm{r})!}=2\cdot \frac{6.5!}{(7-\mathrm{r})(6-\mathrm{r})(5-\mathrm{r})!}\\ \Rightarrow \frac{1}{1}=2\cdot \frac{6}{(7-\mathrm{r})(6-\mathrm{r})}\\ \Rightarrow (7-\mathrm{r})(6-\mathrm{r})=12\\ \Rightarrow 42-13\mathrm{r}+{\mathrm{r}}^2=12\\ \Rightarrow 30-13\mathrm{r}+{\mathrm{r}}^2=0\\ \Rightarrow (10-\mathrm{r})(3-\mathrm{r})=0\\ \Rightarrow (10-\mathrm{r})(3-\mathrm{r})=0\\ \Rightarrow \mathrm{r}=10,3\\ \Rightarrow \mathrm{r}=3[\mathrm{Neglecting}10,\mathrm{because}0<\mathrm{r}\le \mathrm{n}.]\\ \left(\mathrm{ii}\right){}^5{\mathrm{P}}_{\mathrm{r}}{=}^6{\mathrm{P}}_{\mathrm{r}-1}\\ \Rightarrow \frac{5!}{(5-\mathrm{r})!}=\frac{6!}{(6-\mathrm{r}+1)!}\\ \Rightarrow \frac{5!}{(5-\mathrm{r})!}=\frac{6!}{(7-\mathrm{r})!}\\ \Rightarrow \frac{5!}{(5-\mathrm{r})!}=\frac{6.5!}{(7-\mathrm{r})\cdot (6-\mathrm{r})\cdot (5-\mathrm{r})!}\\ \Rightarrow 1=\frac{6}{(7-\mathrm{r})(6-\mathrm{r})}\\ \Rightarrow (7-\mathrm{r})(6-\mathrm{r})=6\\ \Rightarrow 42-13\mathrm{r}+{\mathrm{r}}^2=6\\ \Rightarrow 36-13\mathrm{r}+{\mathrm{r}}^2=0\\ \Rightarrow (9-\mathrm{r})(4-\mathrm{r})=0\\ \Rightarrow \mathrm{r}=9,4\\ \Rightarrow \mathrm{r}=4[\text{Neglecting}9,\text{\hspace{0.33em}because}0<\mathrm{r}\le \mathrm{n}.]\end{array}$

Q.19 How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in ‘EQUATION’}=8\\ \mathrm{Number}\text{\hspace{0.17em}of words using each letter once}={}^8{\mathrm{P}}_8\end{array}$ $\begin{array}{l}=\frac{8!}{(8-8)!}\\ =\frac{8!}{0!}\\ =8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\\ =40320\\ \mathrm{Thus},\text{number of words formed by using letters of ‘Equation’}\\ \text{once is 40320.}\end{array}$

Q.20 How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.

(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in ‘MONDAY’}=6\\ \left(\mathrm{i}\right)\text{If 4 letters are used at a time.}\\ \mathrm{Number}\text{of 4-letter words that can be formed from the letters}\\ \text{of ‘MONDAY’, then}\\ \text{Number of words}={}^6{\mathrm{P}}_4\\ =\frac{6!}{(6-4)!}\\ =\frac{6!}{2!}\\ =6\times 5\times 4\times 3\\ =360\end{array}$ $\begin{array}{l}\mathrm{Thus},\text{number of words formed by using 4 letters is 360.}\\ \left(\mathrm{ii}\right)\mathrm{all}\text{letters are used at a time.}\\ \text{Number of words}={}^6{\mathrm{P}}_6\\ =\frac{6!}{(6-6)!}\\ =\frac{6!}{0!}\\ =6\times 5\times 4\times 3\times 2\times 1\\ =720\\ \left(\mathrm{iii}\right)\mathrm{Number}\text{of vowels}=2\\ \mathrm{Number}\text{of ways to select first letter}\\ ={}^2{\mathrm{P}}_1\\ =\frac{2!}{(2-1)!}\\ =2\\ \mathrm{Number}\text{ of ways to select remaining 5 letters}\\ ={}^5{\mathrm{P}}_5\\ =\frac{5!}{(5-5)!}\\ =5\times 4\times 3\times 2\times 1\\ =120\\ \mathrm{So},\mathrm{total}\text{number of words}=2\times 120\\ =240\end{array}$

Q.21 In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in ‘MISSISSIPPI’}=11\\ \mathrm{Number}\text{of I’s}=4\\ \mathrm{Number}\text{of S’s}=4\\ \mathrm{Number}\text{of P’s}=2\end{array}$ $\begin{array}{l}\mathrm{Number}\text{of permutations}=\frac{11!}{4!4!2!}\\ =\frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{4!4!2!}\\ =\frac{11\times 10\times 9\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 1\times 2\times 1}\\ =34650\\ \mathrm{When}\text{4 I’s come together.}\\ \mathrm{Then},\text{\hspace{0.17em}number of words}=\frac{8!}{4!2!}\\ =\frac{8\times 7\times 6\times 5\times 4!}{4!\times 2\times 1}\\ =840\\ \mathrm{So},\text{required distinct permutations}=34650-840\\ =33810\end{array}$

Q.22 In how many ways can the letters of the word PERMUTATIONS be arranged if the

(i) words start with P and end with S,
(ii) vowels are all together,
(iii) there are always 4 letters between P and S?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in ‘PERMUTATIONS’}=12\\ \mathrm{Number}\text{of T’s}=\text{2}\\ \left(\text{i}\right)\mathrm{When}\text{words start with P and end with S,}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}then number of the words formed}=\frac{10!}{2!}\\ =1814400\\ \left(\mathrm{ii}\right)\mathrm{Vowels}\mathrm{are}\text{all together,}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Number of vowels in PERMUTATIONS}=5\\ \mathrm{If}\text{all vowels are considered as a single letter, then}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}number of letters to be arranged}=12-5+1\\ =8\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}then number of the words formed}=\frac{8!}{2!}\times {}^5{\text{P}}_5\\ =20160\times 120\\ =2419200\\ \left(\mathrm{iii}\right)\text{\hspace{0.33em}If\hspace{0.33em}there are always 4 letters between P and S,\hspace{0.33em}then}\\ \mathrm{number}\text{of letters to be arranged}=12-2\\ =10\\ \mathrm{Number}\text{repeated letter T}=\text{2}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Number of ways to arrange 10 letters}=\frac{10!}{2!}\\ =1814400\\ \begin{array}{cccccccccccc}\text{P}& & & & & \text{S}& & & & & & \end{array}\\ \begin{array}{ccccccc}1& 2& 3& 4& 5& 6& 7\end{array}\\ \mathrm{Total}\text{number of places to shift P and S}=7\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}P and S can be interchange, so total number}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}of ways of arrangement of P and S}=2\times 7\\ \therefore \mathrm{Number}\text{of ways to arrange letters}=14\times 1814400\\ =25401600\end{array}$

Q.23 If ⁿC₈ = ⁿC₂, find ⁿC₂.

Ans

$\mathrm{Given}{,}^{\text{n}}{\text{C}}_{\text{8}}={}^{\text{n}}{\text{C}}_{\text{2}}$ $\begin{array}{l}{\Rightarrow }^{\text{n}}{\text{C}}_{\text{8}}={}^{\text{n}}{\text{C}}_{\text{n-2}}[\because {}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}}]\\ \Rightarrow 8=\mathrm{n}-2\\ \Rightarrow 10=\mathrm{n}\\ \mathrm{So},{}^{\mathrm{n}}{\mathrm{C}}_2={}^{10}{\mathrm{C}}_2\\ =\frac{10!}{2!(10-2)!}\\ =\frac{10\times 9\times 8!}{2!8!}\\ =45\end{array}$

Q.24 Determine n if
(i) ²ⁿC₃ : ⁿC₃ = 12 : 1
(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1

Ans

$\begin{array}{l}\left(\text{i}\right)\frac{{}^{\text{2n}}{\text{C}}_{\text{3}}}{{}^{\text{n}}{\text{C}}_{\text{3}}}=\frac{12}{1}\\ \Rightarrow \frac{\frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}}{\frac{\text{n}!}{3!\left(\text{n}-3\right)!}}=12\\ \Rightarrow \frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}\times \frac{3!\left(\text{n}-3\right)!}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)\left(\text{n}-3\right)!}=12\\ \Rightarrow \frac{2\text{n}\left(2\text{n}-1\right)\left(2\text{n}-2\right)\left(2\text{n}-3\right)!}{\left(2\text{n}-3\right)!}\times \frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=12\\ \Rightarrow 2\text{n}\left(2\text{n}-1\right)2\left(\text{n}-1\right)\times \frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=12\\ \Rightarrow \frac{\left(2\text{n}-1\right)}{\text{n}-2}=3\\ \Rightarrow 2\text{n}-1=3\text{n}-6\\ \Rightarrow 6-1=3\text{n}-2\text{n}\\ \Rightarrow 5=\text{n}\\ \Rightarrow \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}n}=5.\\ \left(\text{ii}\right){}^{\text{2n}}{\text{C}}_{\text{3}}:{}^{\text{n}}{\text{C}}_{\text{3}}=\text{11}:\text{1}\\ \frac{{}^{\text{2n}}{\text{C}}_{\text{3}}}{{}^{\text{n}}{\text{C}}_{\text{3}}}=\frac{11}{1}\\ \Rightarrow \frac{\frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}}{\frac{\text{n}!}{3!\left(\text{n}-3\right)!}}=11\\ \Rightarrow \frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}\times \frac{3!\left(\text{n}-3\right)!}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)\left(\text{n}-3\right)!}=11\\ \Rightarrow \frac{2\text{n}\left(2\text{n}-1\right)\left(2\text{n}-2\right)\left(2\text{n}-3\right)!}{\left(2\text{n}-3\right)!}\times \frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=11\\ \Rightarrow 2\text{n}\left(2\text{n}-1\right)2\left(\text{n}-1\right)\times \frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=11\\ \frac{4\left(2\text{n}-1\right)}{\text{n}-2}=11\\ 8\text{n}-4=11\text{n}-22\\ 22-4=11\text{n}-8\text{n}\\ 18=3\text{n}\\ \Rightarrow \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}n}=6\end{array}$

Q.25 How many chords can be drawn through 21 points on a circle?

Ans

$\begin{array}{l}\mathrm{Number}\text{of points on a circle}=\text{21}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of points used for a chord}=\text{2}\\ \mathrm{Number}\text{of chords made by 21 points}={}^{21}{\mathrm{C}}_2\\ =\frac{21!}{2!(21-2)!}\\ =\frac{21\times 20\times 19!}{2\times 1\times 19!}\\ =210\\ \mathrm{Thus},\text{ number of chords in a circle is 210.}\end{array}$

Q.26 In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Ans

$\begin{array}{l}\mathrm{Total}\text{ number of boys}=\text{5}\\ \text{Total number of girls}=4\\ \mathrm{Number}\text{ of boys in a team}=\text{3}\\ \text{Number of girls in a team}=\text{3}\end{array}$ $\begin{array}{l}\text{Number of ways to select 3 boys}={}^5{\mathrm{C}}_3\\ =\frac{5!}{3!(5-3)!}\\ =\frac{5\times 4\times 3!}{3!\times 2!}\\ =10\\ \text{Number of ways to select 3 girls}={}^4{\mathrm{C}}_3\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}=\frac{4!}{3!(4-3)!}\\ =\frac{4\times 3!}{3!\times 1!}\\ =4\\ \mathrm{Total}\text{\hspace{0.17em}number of ways to select 3}\\ \text{boys and 3 girls}=10\times 4\\ =40\end{array}$

Q.27 Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Ans

$\begin{array}{l}\mathrm{Number}\text{of red balls}=\text{6}\\ \mathrm{Number}\text{of white balls}=\text{5}\\ \mathrm{Number}\text{of blue balls}=\text{5}\\ \mathrm{Number}\text{of ways to select 3 red balls}={}^6{\mathrm{C}}_3\\ =\frac{6!}{3!(6-3)!}\end{array}$ $\begin{array}{l}=\frac{6\times 5\times 4\times 3!}{3!.3!}\\ =20\\ \mathrm{Number}\text{of ways to select 3 white balls}={}^5{\mathrm{C}}_3\\ =\frac{5!}{3!(5-3)!}\\ =\frac{5\times 4\times 3!}{3!.2!}\\ =10\\ \mathrm{Number}\text{of ways to select 3 blue balls}={}^5{\mathrm{C}}_3\\ =\frac{5!}{3!(5-3)!}\\ =\frac{5\times 4\times 3!}{3!.2!}\\ =10\\ \mathrm{Number}\text{\hspace{0.17em}of ways of selecting 9 balls}=20\times 10\times 10\\ =2000\end{array}$

Q.28 Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Ans

$\begin{array}{l}\mathrm{Number}\text{of cards in a deck}=52\\ \mathrm{Number}\text{of cards in each combination}=\text{5}\\ \text{Number of aces in a deck}=4\\ \text{Number of ace in each combination}=\text{1}\\ \text{Number of ways to select 1 ace out of 4}={}^4{\mathrm{C}}_1\end{array}$ $\begin{array}{l}=\frac{4!}{1!(4-1)!}\\ =\frac{4!}{1!3!}\\ =4\\ \text{\hspace{0.17em}No. of ways to select 4 cards out of 48}={}^{48}{\mathrm{C}}_4\\ =\frac{48!}{4!(48-4)!}\\ =\frac{48!}{4!44!}\\ =\frac{48\times 47\times 46\times 45\times 44!}{4\times 3\times 2\times 1\times 44!}\\ =2\times 47\times 46\times 45\\ =194580\\ \text{Number of 5 cards combination}=2\times 47\times 46\times 45\times 4\\ =778320\end{array}$

Q.29 In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Ans

$\begin{array}{l}\mathrm{Total}\text{ number of players}=17\\ \mathrm{Number}\text{of players who can bowl}=\text{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of players in cricket team}=11\\ \mathrm{Number}\text{of bowlers in cricket team}=4\end{array}$ $\begin{array}{l}\mathrm{Number}\text{\hspace{0.17em}of players who can not bowl}=17-5\\ =12\\ \mathrm{Number}\text{ of ways to select 11 players}={}^{12}{\mathrm{C}}_7\times {}^5{\mathrm{C}}_4\\ =\frac{12!}{7!(12-7)!}\times \frac{5!}{4!(5-4)!}\\ =\frac{12!}{7!\times 5!}\times \frac{5!}{4!1!}\\ =\frac{12\times 11\times 10\times 9\times 8\times \bcancel{7!}}{\bcancel{7!}\times \bcancel{5!}}\times \frac{\bcancel{5!}}{4!}\\ =3960\end{array}$

Q.30 A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Ans

$\begin{array}{l}\mathrm{Number}\text{of black balls in a bag}=5\\ \mathrm{Number}\text{of red balls in a bag}=6\\ \mathrm{Number}\text{of ways to select 2 black and 3 red balls}\\ ={}^5{\text{C}}_2\times {}^6{\text{C}}_3\\ =\frac{5!}{2!\left(5-2\right)!}\times \frac{6!}{3!\left(6-3\right)!}\\ =\frac{5!}{2!\times 3!}\times \frac{6!}{3!\times 3!}\\ =\frac{5\times 4\times \bcancel{3!}}{2!\times \bcancel{3!}}\times \frac{\bcancel{6}\times 5\times 4\times \bcancel{3!}}{\bcancel{3!}\times \bcancel{3!}}\\ =10\times 5\times 4\\ =200\end{array}$

Q.31 In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Ans

$\begin{array}{l}\mathrm{Available}\text{courses for students}=9\\ \mathrm{Number}\text{of specific courses}=2\\ \mathrm{Courses}\text{to be chosen by students}=5\\ \mathrm{Not}\text{specific courses for students}=5-2\\ =3\\ \mathrm{Number}\text{of ways to select 3 remaining courses}\\ ={\text{\hspace{0.17em}\hspace{0.17em}}}^7{\mathrm{C}}_3\\ =\frac{7!}{3!\times (7-3)!}[\because {}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}]\\ =\frac{7!}{3!\times 4!}\\ =\frac{7\times 6\times 5\times 4!}{3!\times 4!}\\ =35\\ \mathrm{Thus},\text{a student can choose a programme of 5 courses in 35 ways.}\end{array}$

Q.32 How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in the word ‘DAUGHTER’}=8\\ \mathrm{Number}\text{of vowels in the word ‘DAUGHTER’}=3\\ \mathrm{Number}\text{of consonants in the word ‘DAUGHTER’}=5\\ \mathrm{Ways}\text{to select 2 vowels and 3 consonants}={}^3{\mathrm{C}}_2\times {}^5{\mathrm{C}}_3\\ =\frac{3!}{2!(3-2)!}\times \frac{5!}{3!(5-3)!}\\ =\frac{1}{2!\times 1!}\times \frac{5!}{2!}\\ =\frac{120}{4}\\ =30\\ \mathrm{Number}\text{of ways to arrange 2 vowels and 3 consonants}\\ =30\times 5!\end{array}$ $\begin{array}{l}=30\times 120\\ =3600\\ \mathrm{Thus},\text{required number of different words is 3600.}\end{array}$

Q.33 How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letter in the word ‘EQUATION’}=8\\ \mathrm{Number}\text{of vowel in the word ‘EQUATION’}=5\\ \mathrm{Number}\text{of ways to arrange vowels}={}^5{\text{P}}_5\\ =5!\\ =120\\ \mathrm{Number}\text{\hspace{0.33em}of ways to arrange 3 consonants}={}^3{\text{P}}_3\\ =3!\\ =6\\ \mathrm{Since},\text{vowels and consonants will occur together, both \hspace{0.33em}vowels}\left(\mathrm{EUAIO}\right)\text{\hspace{0.33em}and consonants}\\ \left(\mathrm{QTN}\right)\text{can be considered \hspace{0.33em}two objects. So, number of ways to arrange two objects}\\ ={}^2{\text{P}}_2=2\\ \mathrm{Thus},\text{number of words formed by using}=120\times 6\times 2\\ =1440\end{array}$

Q.34 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) atleast 3 girls ?
(iii) atmost 3 girls ?

Ans

$\begin{array}{l}\mathrm{Number}\text{of members in a commitee}=7\\ \mathrm{Number}\text{of total boys for commitee}=9\\ \mathrm{Number}\text{of total girls for commitee}=4\\ \left(\mathrm{i}\right)\mathrm{If}\text{ there are exactly 3 girls in commitee.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} Number of ways to select 3 girls}={}^4{\mathrm{C}}_3\\ =\frac{4!}{3!(4-3)!}\\ =4\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} Number of ways to select 4 boys}={}^9{\mathrm{C}}_4\\ =\frac{9!}{4!(9-4)!}\end{array}$ $\begin{array}{l}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\\ =126\\ \mathrm{Number}\text{of ways to form a commitee}=4\times 126\\ =504\\ \left(\mathrm{ii}\right)\mathrm{If}\text{ there are atleast 3 girls in commitee.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of ways to form commitee}={}^4{\mathrm{C}}_3{\times }^9{\mathrm{C}}_4{+}^4{\mathrm{C}}_4{\times }^9{\mathrm{C}}_3\\ =\frac{4!}{3!(4-3)!}\times \frac{9!}{4!(9-4)!}\\ +\frac{4!}{4!(4-4)!}\times \frac{9!}{3!(9-3)!}\\ =\frac{4!}{3!1!}\times \frac{9!}{4!5!}+\frac{4!}{4!0!}\times \frac{9!}{3!6!}\\ =4\times 126+1\times 84\\ =504+84\\ =588\\ \left(\mathrm{iii}\right)\mathrm{If}\text{ there are atmost 3 girls in commitee.}\\ \text{There are different ways to complete 7 persons in commitee.}\\ \text{When 3 girls and 4 boys are selected}={}^4{\mathrm{C}}_3{\times }^9{\mathrm{C}}_4\\ =\frac{4!}{3!(4-3)!}\times \frac{9!}{4!(9-4)!}\\ =4\times 126=504\\ \text{When 2 girls and 5 boys are selected}={}^4{\mathrm{C}}_2\times {}^9{\mathrm{C}}_5\\ =\frac{4!}{2!(4-2)!}\times \frac{9!}{5!(9-5)!}\\ =6\times 126\end{array}$ \begin{array}{l}\mathrm{}\end{array} $\begin{array}{l}=756\\ \text{When 1 girl and 6 boys are selected}={}^4{\mathrm{C}}_1\times {}^9{\mathrm{C}}_6\\ =\frac{4!}{1!(4-1)!}\times \frac{9!}{6!(9-6)!}\\ =4\times 84\\ =336\\ \text{When no girl and 6 boys are selected}={}^4{\mathrm{C}}_0\times {}^9{\mathrm{C}}_7\\ =\frac{4!}{0!(4-0)!}\times \frac{9!}{7!(9-7)!}\\ =1\times 36\\ =36\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of ways to form commitee}={}^4{\mathrm{C}}_3{\times }^9{\mathrm{C}}_4+{}^4{\mathrm{C}}_2{\times }^9{\mathrm{C}}_5\\ +{}^4{\mathrm{C}}_1\times {}^9{\mathrm{C}}_6+{}^4{\mathrm{C}}_0\times {}^9{\mathrm{C}}_7\\ =504+756+336+36\\ =1632\end{array}$

Q.35 If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in word ‘EXAMINATION’}=11\\ \mathrm{Number}\text{of letter A in word ‘EXAMINATION’}=2\\ \mathrm{Number}\text{of letter I in word ‘EXAMINATION’}=2\\ \mathrm{Number}\text{of letter N in word ‘EXAMINATION’}=2\\ \mathrm{The}\text{words starting with A will come first in dictionary from}\\ \text{the words starting with E.}\end{array}$ $\begin{array}{l}\text{So, the number of words starting with letter A}\\ =\frac{10!}{2!2!}\\ =\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2!}{2!2!}\\ =5\times 9\times 8\times 7\times 6\times 5\times 4\times 3\\ =907200\\ \mathrm{Thus},\text{there are 907200 words listed in dictionary formed by}\\ \text{the letters of ‘EXAMINATION’ before the first word starting with}\\ \text{letter E.}\end{array}$

Q.36 How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?

Ans

$\begin{array}{l}\mathrm{Since},\text{a number will be divisible by 10 if its last digit is 0.}\\ \text{Number of given digits}=6\\ \mathrm{Last}\text{digit of number formed is fixed i.e., 0.}\\ \text{Since, no digit is repeated.}\end{array}$

$\begin{array}{l}\text{Number of 6-digit numbers formed}=\mathrm{ways}\text{to fill remaining places}\\ \text{\hspace{0.17em}of 6-digit numbers}\\ =5\times 4\times 3\times 2\times 1\\ =120\\ \mathrm{Thus},\text{the number of 6-digit numbers formed by given digits}\\ \text{and divisible by 10 is 120.}\end{array}$