# NCERT SOLUTIONS FOR CLASS 11 MATHS CHAPTER 7 / PERMUTATIONS AND COMBINATIONS EX 7.4 /

Question 1. lf nC8 = nC2, find nC2.

Solution.

We have, nC8 = nC2

Question 2. Determine n if

• (i) 2nC3nC3 =12 : 1
• (ii) 2nC3nC3= 11 : 1

Solution.

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

Solution.

A chord is formed by joining two points on a circle.

∴ Required number of chords = 2nC2 $=\frac { 21! }{ 2!19! } =\frac { 21\times 20 }{ 2 } =210$

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

Solution.

3 boys can be selected from 5 boys in 5C3 ways & 3 girls can be selected from 4 girls in 4C3 ways.

∴ Required number of ways of team selection = 5C3 x 4C3 = $\frac { 5! }{ 2!3! } \times \frac { 4! }{ 3!1! }$ $=\frac { 5\times 4 }{ 2 } \times 4=40$

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.

No. of ways of selecting 3 red balls =6C3

No. of ways of selecting 3 white balls = 5C3

No. of ways of selecting 3 blue balls = 5C3

∴ Required no. of ways of selecting 9 balls

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

Solution.

Total no. of cards = 52

No. of ace cards = 4

No. of non-ace cards = 48

∴ One ace card out of 4 can be selected in 4C1 ways.

Remaining 4 cards out of 48 cards can be selected in 48C4ways.

∴ Required no. of ways of selecting 5 cards

Question 7. In Kbw 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.

Total players = 17, No. of bowlers = 5,

No. of non-bowlers = 12

No. of ways of selecting 4 bowlers = 5C4

No. of ways of selecting 7 non-bowlers = 12C7

∴ Required no. of ways of selecting a cricket team

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.

No. of ways of selecting 2 black balls = 5C2

No. of ways of selecting 3 red balls = 6C3

∴ Required no. of ways of selecting 2 black & 3 red balls = 5C2 x 6C3 $=\frac { 5! }{ 2!3! } \times \frac { 6! }{ 3!3! } =\frac { 5\times 4 }{ 2 } \times \frac { 6\times 5\times 4 }{ 3\times 2 } =200$

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.

Total no. of courses = 9

No. of compulsory courses = 2

So, the student will choose 3 courses out of 7 courses [non-compulsory courses].

∴ Required no. of ways a student can choose a programme = 7C3 = $\frac { 7! }{ 3!4! } =\frac { 7\times 6\times 5 }{ 6 } =35$

#### Verb Notes

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Phasellus cursus rutrum est nec suscipit. Ut et ultrices nisi. Vivamus id nisl ligula. Nulla sed iaculis ipsum.

Company Name