NCERT Solutions for Class 11 Maths chapter-7 Permutation And Combination Exercise 7.4

NCERT Solutions for Class 11 Maths chapter-7 Permutation And Combination Exercise 7.4

NCERT Solutions for Class-11 Maths Chapter-7 Permutation and Combination

NCERT Solutions For Class 11 Maths Chapter-7 Permutation and Combination prepared by the expert of Physics Wallah score more with Physics Wallah NCERT Class 11 maths solutions. You can download solution of all chapters from Physics Wallah NCERT solutions of class 11.

NCERT Solutions for Class-11 Maths Exercise 7.4

Question 1. if ⁿ C ₈ = ⁿ C ₂ ,find ⁿ C ₂

Solution :
Given: ⁿ C ₈ = ⁿ C ₂ ,find ⁿ C ₂

Since, we know that ⁿ C _a = ⁿ C _b ⇒a=b or m=a+b

That is,

ⁿ C _{2 =} ¹⁰ C ₂ =10! / 2!(10−2)! = 10! / 2!8!=10×9×8!  /  2×1×8!= 45

Hence, ⁿ C 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) Given: ²ⁿ C ₃ : ⁿ C ₃ =12:1

(ii) Given: ²ⁿ C ₃ : ⁿ C ₃ = 11:1

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

Solution :
There are 21 points on the circumference of a circle. Since one and only one chord can be drawn by joining 2 distinct points, therefore the required number of chords given by

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

Solution :
There are 5 boys and 4 girls and 3 out of 5 boys and 3 out of 4 girls have to be selected.

Number of ways of selection =

Question 5. Find the number 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 : Given: Total number of balls: 6 red balls, 5 white balls and 5 blue balls.

3 balls of each colour should be selected making a total of 9 balls.

Therefore,

Out of 6 red balls, 3 balls can be selected in ⁶ C ₃

ways

Out of 5 white balls, 3 balls can be selected in ⁵ C ₃

ways

Out of 5 blue balls, 3 balls can be selected in ⁵ C ₃

ways

Therefore, by multiplication principle, number of ways of selecting 9 balls is =

= 20 × 10  × 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 :
There are 4 aces and 48 other cards in a deck of 52 cards. We have to select 1 ace out of 4 aces and 4 other cards out of 48 other cards.

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 : There are 5 bowlers out of 17 players.


A cricket team of 11 is selected with exactly 4 bowlers.


Thus, 4 bowlers can be selected in ⁵ C4
ways and out of remaining 12 players, 7 players are selected in ¹² C ₇

ways.

Thus, by multiplication principle, required number of ways of selecting 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 :
There are 5 black and 6 red balls. We have to select 2 black balls out of 5 black balls and 3 red balls out of 6 red balls.

⇒ Number of ways of selection

=  200

Question 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 and number of courses to be selected are 5 in which 2 specific courses are compulsory. 3 courses out of remaining 7 courses have to be selected.

⇒ Number of ways of selection = ⁷ c ₃

= = 35