NCERT Solutions Class 11th Maths Chapter – 7 Permutations and Combinations Exercise 7.4
Textbook | NCERT |
class | Class – 11th |
Subject | Mathematics |
Chapter | Chapter – 7 |
Chapter Name | Permutations and Combinations |
grade | Class 11th Maths solution |
Medium | English |
Source | last doubt |
NCERT Solutions Class 11th Maths Chapter – 7 Permutations and Combinations Exercise 7.4
?Chapter – 7?
✍Permutations and Combinations✍
?Exercise 7.4?
1. If nC8 = nC2, find nC2.
Solution:
2. Determine n if
(i) 2nC3:nC3 = 12: 1
(ii) 2nC3: nC3 = 11: 1
Solution:
Simplifying and computing
⇒ 4 × (2n – 1) = 12 × (n – 2)
⇒ 8n – 4 = 12n – 24
⇒ 12n – 8n = 24 – 4
⇒ 4n = 20
∴ n = 5
⇒ 11n – 8n = 22 – 4
⇒ 3n = 18
∴ n = 6
3. How many chords can be drawn through 21 points on a circle?
Solution:
Given 21 points on a circle
We know that we require two points on the circle to draw a chord
∴ Number of chords is are
⇒ 5C3 × 4C3 =
∴ Total number of chords can be drawn are 210
4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Solution:
Given 5 boys and 4 girls are in total
We can select 3 boys from 5 boys in 5C3 ways
Similarly, we can select 3 boys from 54 girls in 4C3 ways
∴ Number of ways a team of 3 boys and 3 girls can be selected is 5C3 × 4C3
⇒ 5C3 × 4C3 = 10 × 4 = 40
∴ Number of ways a team of 3 boys and 3 girls can be selected is 5C3 × 4C3 = 40 ways
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:
Given 6 red balls, 5 white balls and 5 blue balls
We can select 3 red balls from 6 red balls in 6C3 ways
Similarly, we can select 3 white balls from 5 white balls in 5C3 ways
Similarly, we can select 3 blue balls from 5 blue balls in 5C3 ways
∴ Number of ways of selecting 9 balls is 6C3 ×5C3 × 5C3
∴ 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 is 6C3 ×5C3 × 5C3 = 2000
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:
Given a deck of 52 cards
There are 4 Ace cards in a deck of 52 cards.
According to question, we need to select 1 Ace card out the 4 Ace cards
∴ Number of ways to select 1 Ace from 4 Ace cards is 4C1
⇒ More 4 cards are to be selected now from 48 cards (52 cards – 4 Ace cards)
∴ Number of ways to select 4 cards from 48 cards is 48C4
∴ Number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination 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:
Given 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers
There are 5 players how bowl, and we can require 4 bowlers in a team of 11
∴ Number of ways in which bowlers can be selected are: 5C4
Now other players left are = 17 – 5(bowlers) = 12
Since we need 11 players in a team and already 4 bowlers are selected, we need to select 7 more players from 12.
∴ Number of ways we can select these players are: 12C7
∴ Total number of combinations possible are: 5C4 × 12C7
∴ Number of ways we can select a team of 11 players where 4 players are bowlers from 17 players are 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:
Given a bag contains 5 black and 6 red balls
Number of ways we can select 2 black balls from 5 black balls are 5C2
Number of ways we can select 3 red balls from 6 red balls are 6C3
Number of ways 2 black and 3 red balls can be selected are 5C2× 6C3
∴ Number of ways in which 2 black and 3 red balls can be selected from 5 black and 6 red balls are 200
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:
Given 9 courses are available and 2 specific courses are compulsory for every student
Here 2 courses are compulsory out of 9 courses, so a student need to select 5 – 2 = 3 courses
∴ Number of ways in which 3 ways can be selected from 9 – 2(compulsory courses) = 7 are 7C3
∴ Number of ways a student selects 5 courses from 9 courses where 2 specific courses are compulsory are: 35