Chapter 6: Permutations and Combinations

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

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, $n=5$.

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, $n=6$.

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

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

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

$\Rightarrow 210$

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

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

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

$\Rightarrow 40$

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 –

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

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

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

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 –

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

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

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

$\Rightarrow 778320$

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 –

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

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

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

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

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

$\Rightarrow 200$

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 Solution Class 11 Maths of Chapter 1 All Exercises