**If**^{n}C_{8 }=^{n}C_{2}. find^{n}C_{2}

Solution:

We know that, ^{n}C_{a} = ^{n}C_{b} then a = b or n = a + b

Therefore,

^{n}C_{8 }= ^{n}C_{2}, since 8≠2 , we have 8 + 2 = 10

^{n}C_{2} = ^{10}C_{2} = ^{10!}/_{2!(10-2)!} = ^{10!}/_{2!8!} = ^{10x9x8}/_{2x1x8} = 45

**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

(ii) ^{2n}C_{3} : ^{n}C_{3} = 11 : 1

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

solution:

We know that, to draw one chord on 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, there will be as many chords as there are combinations of 21 points taken 2 at a time.

Thus, required number of chords = ^{21}C_{2} = ^{21!}/_{2!(21-2)!} = ^{21!}/_{21!9!} = ^{21×20}/_{2} = 210

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

Solution:

It is given that, a team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls.

3 boys can be selected from 5 boys in ^{5}C_{3 } ways.

3 girls can be selected from 4 girls in ^{4}C_{3} ways.

Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected = ^{5}C_{3} x ^{4}C_{3 } = ^{5!}/_{3!2!} x ^{4!}/_{3!1!}

= ^{5x4x3!}/_{3!} x ^{4×3!}/_{3!}

= 10 x 4

= 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 of each colour.**

Solution:

It is given that, there are a total of 6 red balls, 5 white balls, and 5 blue balls. 9 balls have to be selected in such a way that each selection consists of 3 balls of each colour.

3 balls can be selected from 6 red balls in ^{6}C_{3} ways.

3 balls can be selected from 5 white balls in ^{5}C_{3 }ways.

3 balls can be selected from 5 blue balls in ^{5}C_{3 }ways.

Thus, by multiplication principle, required number of ways of selecting 9 balls

= ^{6}C_{3} x ^{5}C_{3} x ^{5}C_{3}

= ^{6!}/_{3!3!} x ^{5!}/_{3!2!} x ^{5!}/_{3!2!}

= ^{6x5x4x3x2x1}/_{3!x3x2} x ^{5x4x3x2x1}/_{3!x2} x ^{5x4x3x2x1}/_{3!x2}

= 20 x 10 x 10

= 2000

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

solution:

IT is given that, in a deck of 52 cards, there are 4 aces.

A combination of 5 cards have to be made in which there is exactly one ace.

Then, one ace can be selected in ^{4}C_{1 } ways and the remaining 4 cards can be selected out of the 48 cards in ^{48}C_{4} ways.

Thus, by multiplication principle, required number of 5 card combinations

^{48}C_{4} x ^{4}C_{1} = ^{48!}/_{4!44!} x ^{4!}/_{1!3!}

= ^{48x47x46x45}/_{4x3x2x1} x 4

= 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?**

Solution:

It is given that out of 17 players, 5 players are bowlers.

A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers.

4 bowlers can be selected in ^{5}C_{4} ways and the remaining 7 players can be selected out of the 12 players in ^{12}C_{7} ways.

Thus, by multiplication principle, required number of ways of selecting cricket team = ^{5}C_{4} x ^{12}C_{7 }= ^{5!}/_{4!1!} x ^{12!}/_{7!5!} = 5 x ^{12x11x10x9x8}/_{5x4x3x2x1} = 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.**

solution:

We have 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} x ^{6}C_{3} = ^{5!}/_{2!3!} x ^{6!}/_{3!3!} = ^{5×4}/_{2} x ^{6x5x4}/_{3x2x1} = 10 x20 = 200

**9: In how many ways can a student choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?**

solution:

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 the required number of ways of choosing the programme

= ^{7}C_{3} = ^{7!}/_{3!4!} = ^{7x6x5x4!}/_{3x2x1x4} = 35