NCERT Solutions for Class 11 Maths Chapter 7: Permutations and Combinations - Exercise 7.4

Exercise-7.4

1. If $^{n}{{C}_{8}}{{=}^{n}}{{C}_{2}}$, find $^{n}{{C}_{2}}$.

Ans: Since, we know that \[^{n}{{C}_{a}}{{=}^{n}}{{C}_{b}}\]

\[\Rightarrow a=b\]

Or, \[n=a+b\]

Here n=8+2=10

Therefore,

$^{10}{{C}_{2}}=\dfrac{10!}{2!(10-2)!}$

\[\Rightarrow \dfrac{10!}{2!8!}=\dfrac{10\times 9\times 8!}{2\times 1\times 8!}\]

Hence, $^{n}{{C}_{2}}=45$.

2. Determine $n$ if

(i) $^{2n}{{C}_{3}}:{{\text{ }}^{n}}{{C}_{3}}=12:1$

Ans: $\dfrac{^{2n}{{C}_{3}}}{^{n}{{C}_{3}}}=\dfrac{12}{1}$

\[\Rightarrow \dfrac{(2n)!}{3!(2n-3)!}\times \dfrac{3!(n-3)!}{n!}=\dfrac{12}{1}\]

\[\Rightarrow \dfrac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!}\times \dfrac{(n-3)!}{n(n-1)(n-2)(n-3)!}=12\]

Therefore,

\[\Rightarrow \dfrac{2(2n-1)(2n-2)}{(n-1)(n-2)}=12\]

\[\Rightarrow \dfrac{4(2n-1)(n-1)}{(n-1)(n-2)}=12\]

\[\Rightarrow 2n-1=3(n-2)\]

\[\Rightarrow 3n-2n=-1+6\]

Hence, $n=5$.

(ii) $^{2n}{{C}_{3}}:{{\text{ }}^{n}}{{C}_{3}}=11:1$

Ans: \[\Rightarrow \dfrac{(2n)!}{3!(2n-3)!}\times \dfrac{3!(n-3)!}{n!}=\dfrac{11}{1}\]

\[\Rightarrow \dfrac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!}\times \dfrac{(n-3)!}{n(n-1)(n-2)(n-3)!}=11\]

\[\Rightarrow \dfrac{2(2n-1)(2n-2)}{(n-1)(n-2)}=11\]

Therefore,

\[\Rightarrow \dfrac{(2n-1)}{(n-2)}=\dfrac{11}{4}\]

\[\Rightarrow 8n-4=11n-22\]

\[\Rightarrow 3n=18\]

Hence, $n=6$.

3. How many chords can be drawn through $21$ points on a circle?

Ans: For drawing one chord a circle, only $2$ points are required.

To know the number of chords that can be drawn through the given $21$ points on a circle, the number of combinations have to be counted.

Therefore, the chords can be drawn through $21$ points taken $2$ as equal to each chord.

Thus, required number of chords

$^{21}{{C}_{2}}=\dfrac{21!}{2!(21-2)!}$

$\Rightarrow \dfrac{21!}{2!19!}=\dfrac{21\times 20}{2}$

$\Rightarrow 210$

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

Ans: The team which is selected is of $3$ boys and $3$ girls from $5$ boys and $4$ girls.

The team of $3$ boys can be selected from $5$ boys in $^{5}{{C}_{3}}$ different ways.

The team of $3$ girls can be selected from $4$ girls in $^{4}{{C}_{3}}$ different ways.

Therefore, by multiplication principle, number of ways in which a team of $3$ boys and $3$ girls can be selected

${{\Rightarrow }^{5}}{{C}_{3}}{{\times }^{4}}{{C}_{3}}=\dfrac{5!}{3!2!}\times \dfrac{4!}{3!1!}$

$\Rightarrow \dfrac{5\times 4\times 3!}{3!\times 2}\times \dfrac{4\times 3!}{3!}=10\times 4$

$\Rightarrow 40$

5. 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 each colour.

Ans: The number of balls from which we have to select $6$ red balls, $5$ white balls, and $4$ blue balls is $9$ balls. These balls have to be selected in such a manner that each selection of balls consists of $9$ balls from each of the colour.

Here, given we have :

$3$ balls which can be opted from the $6$ red balls in $^{6}{{C}_{3}}$ different ways.

$3$ balls which can be opted from the $5$ white balls in $^{5}{{C}_{3}}$ different ways.

$3$ balls which can be opted from the $5$ blue balls in $^{5}{{C}_{3}}$ different ways.

Thus, after applying the multiplication principle, required number of ways of selecting the $9$ balls will be –

${{\Rightarrow }^{6}}{{C}_{3}}{{\times }^{5}}{{C}_{3}}{{\times }^{5}}{{C}_{3}}=\dfrac{6!}{3!3!}\times \dfrac{5!}{3!2!}\times \dfrac{5!}{3!2!}$

$\Rightarrow \dfrac{6\times 5\times 4\times 3!}{3!3\times 2}\times \dfrac{5\times 4\times 3!}{3!2\times 1}\times \dfrac{5\times 4\times 3!}{3!2\times 1}$

$\Rightarrow 20\times 10\times 10=2000$

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

Ans: We have a deck of $52$ cards, which contains $4$ aces. When we make a combination, $5$ cards should be made in such a manner that there is exactly one ace.

Now, $1$ ace can be opted in $^{4}{{C}_{3}}$ different ways and the cards which are left, out of them $4$ cards can be opted out of the $48$ cards in $^{48}{{C}_{4}}$ different ways.

Therefore, after applying multiplication principle, the required number of $5$ card combinations will be –

${{\Rightarrow }^{48}}{{C}_{4}}{{\times }^{4}}{{C}_{1}}=\dfrac{48!}{4!44!}\times \dfrac{4!}{1!3!}$

$\Rightarrow \dfrac{48\times 47\times 46\times 45\times 44!}{44!4!\times 3\times 2\times 1}\times 4!$

$\Rightarrow \dfrac{48\times 47\times 46\times 45}{3\times 2\times 1}$

$\Rightarrow 778320$

7. 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: The number of players out of which we have to select is $17$ players and only $5$ players from them are bowlers.

A cricket team of $11$ players need to be opted in such a way that there remains exactly $4$ players which are bowlers.

$4$ bowlers can be opted in $^{5}{{C}_{4}}$ different ways whereas out of the remaining $7$ players can be opted out from the $12$ players in $^{12}{{C}_{7}}$ different ways.

Therefore, after applying multiplication principle, the required number of ways for selecting the cricket team will be –

$^{5}{{C}_{4}}{{\times }^{12}}{{C}_{7}}=\dfrac{5!}{4!1!}\times \dfrac{12!}{7!5!}$

$\Rightarrow 5!\times \dfrac{12\times 11\times 10\times 9\times 8}{5!\times 4\times 3\times 2\times 1}=3960$

8. 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: There are $5$ black and $6$ red balls in the bag. $2$ black balls can be selected out of $5$ black balls in $^{5}{{C}_{2}}$ ways and $3$ red balls can be selected out of $6$ red balls in $^{6}{{C}_{3}}$ ways.

Thus, by multiplication principle, required number of ways of selecting $2$ black and $3$ red balls

$^{5}{{C}_{2}}{{\times }^{6}}{{C}_{3}}=\dfrac{5!}{2!3!}\times \dfrac{6!}{3!3!}$

$\Rightarrow \dfrac{5\times 4}{2}\times \dfrac{6\times 5\times 4}{3\times 2\times 1}=10\times 20$

$\Rightarrow 200$

9. 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: There are $9$ courses available out of which, $2$ specific courses are compulsory for every student.

Therefore, every student has to choose $3$ courses out of the remaining $7$ courses. This can be chosen in $^{7}{{C}_{3}}$ ways.

Thus, required number of ways of choosing the programme is 

$^{7}{{C}_{3}}=\dfrac{7!}{3!4!}$

$\Rightarrow \dfrac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}=35$

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

NCERT Solution Class 11 Maths of Chapter 7 Exercise

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Exercise 7.4

Opting for the NCERT solutions for Ex 7.4 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.4 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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