11th NCERT Permutations And Combinations
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## Exercise 7.4 Questions 9

Question (1)

If,$n{C_8} = n{C_{2,}}find\;n{C_2}$

Solution

$n{C_8} = n{C_{2,}}$ $\Rightarrow 8 = 2\;which\;is\;not\;possible$ $or\;n = 2 + 8 = 10$ $n{C_{2,}} = 10{C_2}$ $= \frac{{10 \times 9}}{2}$ $= 45$

Question (2)

Determine n if (i) $2n{C_3}:n{C_3} = 12:1$ (ii) $2n{C_3}:n{C_3} = 11:1$

Solution

(i) $2n{C_3}:n{C_3} = 12:1$ $\Rightarrow \frac{{2n{C_3}}}{{n{C_3}}} = \frac{{12}}{1}$ $\Rightarrow \frac{{\frac{{2n\left( {2n - 1} \right)\left( {2n - 2} \right)}}{{3 \times 2}}}}{{\frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3 \times 2}}}} = \frac{{12}}{1}$ $\Rightarrow \frac{{2n\left( {2n - 1} \right)2\left( {n - 1} \right)}}{{n\left( {n - 1} \right)\left( {n - 2} \right)}} = \frac{{12}}{1}$ $\Rightarrow \frac{{2n - 1}}{{n - 2}} = \frac{{12}}{4} = 3$ $\Rightarrow 2n - 1 = 3n - 6$ $\Rightarrow n = 5$ (ii) $2n{C_3}:n{C_3} = 11:1$ $\Rightarrow \frac{{2n{C_3}}}{{n{C_3}}} = \frac{{11}}{1}$ $\Rightarrow \frac{{\frac{{2n\left( {2n - 1} \right)\left( {2n - 2} \right)}}{{3 \times 2}}}}{{\frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3 \times 2}}}} = \frac{{11}}{1}$ $\Rightarrow \frac{{2n\left( {2n - 1} \right)2\left( {n - 1} \right)}}{{n\left( {n - 1} \right)\left( {n - 2} \right)}} = \frac{{11}}{1}$ $\Rightarrow \frac{{2n - 1}}{{n - 2}} = \frac{{11}}{4}$ $\Rightarrow 8n - 4 = 11n - 22$ $\Rightarrow 18 = 3n$ $\Rightarrow n = 6$

Question (3)

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

Solution

There are 21 points on the circle. To form a chord two points are required.
number of chords that can be drawn
$= 21{C_2}$ $= \frac{{21 \times 20}}{2}$ $= 210$

Question (4)

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

Solution

We have to select the team of 3 boys and 3 girls from 5 boys and 4 girls.
The selection of 3 boys from 5 boys can be done in 5C3 different ways.
The selection of 3 girls from 4 girls can be done in 4C3 different ways.
Number of ways the team is selected
$= 5{C_3} \times 4{C_3}$ $= \frac{{5 \times 4 \times 3}}{{3 \times 2}} \times \frac{{4 \times 3 \times 2}}{{3 \times 2}}$ $= 10 \times 4$ $= 40$

Question (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 of each colour.

Solution

We have to select 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
So we have to select 3 red from 6 red balls , it can be done in 6C3 ways.
So we have to select 3 white from 5 white balls , it can be done in 5C3 ways.
So we have to select 3 blue from 5 blue balls , it can be done in 5C3 ways.
Number of ways of selecting balls
$= 6{C_3} \times 5{C_3} \times 5{C_3}$ $= \frac{{6 \times 5 \times 4}}{{3 \times 2}} \times \frac{{5 \times 4 \times 3}}{{3 \times 2}} \times \frac{{5 \times 4 \times 3}}{{3 \times 2}}$ $= 20 \times 10 \times 10$ $= 2000$

Question (6)

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

Solution

In a combination of 5 cards it requires exactly 1 ace.
So out ot 4 aces we have to select 1 ace, it can be done in 4C1 ways.
Now out of 48 (52 - 4) cards we have to selct 4 cards, it can be done in 48C4 ways.
So the number of combinations of selections of 5 cards
$= 4{C_1} \times 48{C_4}$ $= 4 \times \frac{{48 \times 47 \times 46 \times 45}}{{4 \times 3 \times 2}}$ $= 778320$

Question (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?

Solution

In 17 players , 5 are bowlers. In a team of 11 must include exactly 4 bowlers.
The selection of bawlers can be done in 5C4 ways.
Now (17-5) from 12 players we have to select 7 players to form a team of 11 players.
It can be done by 12 C7 different ways.
The number of ways of selecting the team
$= 5{C_4} \times 12{C_7}$ $= \frac{{5 \times 4 \times 3 \times 2}}{{4 \times 3 \times 2}} \times \frac{{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6}}{{7 \times 6 \times 5 \times 4 \times 3 \times 2}}$ $= 5 \times 792$ $= 3960$

Question (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.

Solution

The selection of 2 black balls from 5 black balls is done by 5C2 ways.
The selection of 3 red balls from 6 red balls is done by 6C3 ways.
The number of ways of selection
$5{C_2} \times 6{C_3}$ $= \frac{{5 \times 4}}{{2 \times 1}} \times \frac{{6 \times 5 \times 4}}{{3 \times 2}}$ $= 10 \times 20$ $= 200$

Question (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?

Solution

A student has to choose the programme of 5 courses in which 2 specific courses are compulsasory. So out of 7 ( 9 - 2 ) he has to select 3 courses,
it can be done in 7C3 ways.
Number of ways student can choose the programme
$= 7{C_3}$ $\frac{{7 \times 6 \times 5}}{{3 \times 2}}$ $= 35$