Solve all your doubts with Teachoo Black (new monthly pack available now!)
Example 15 - Chapter 7 Class 11 Permutations and Combinations (Term 2)
Last updated at Dec. 8, 2016 by Teachoo
Are you in school? Do you love Teachoo?
We would love to talk to you! Please fill this form so that we can contact you
Examples
Example 1
Example 2
Example 3 Important
Example 4 Important
Example 5
Example 6 (i)
Example 6 (ii)
Example 7
Example 8 Important
Example 9 Important
Example 10
Example 11 Important
Example 12 (i) Important
Example 12 (ii)
Example 13
Example 14 Important
Example 15 You are here
Example 16 Important
Example 17
Example 18
Example 19 Important
Example 20 Important
Example 21 Important
Example 22 Important
Example 23 Important
Example 24 Important
Chapter 7 Class 11 Permutations and Combinations
Serial order wise
- Ex 7.1
- Ex 7.2
- Ex 7.3
- Ex 7.4
-
Examples
- Miscellaneous
Transcript
Example 15 In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable ? Total number of discs = 4 red + 3yellow + 2 green n = 9. Since color are repeating so we use this formula π!/π1!π2!π3! Here n = 9 Out of 9 discs, 4 red, 3 yellow and 2 green & p1 = 4, p2 = 3, p3 = 2 The number of arrangement = π!/π1!π2!π3! = 9!/(4! 3! 2!) = (9 Γ 8 Γ 7 Γ 6 Γ 5 Γ 4!)/(4! Γ (3 Γ 2 Γ 1)(2 Γ 1)) = (9 Γ 8 Γ 7 Γ 6 Γ 5)/(3 Γ 2 Γ 2) = 1260. Hence, total number of arrangements is 1260
