Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Example 12 (i) - Chapter 6 Class 11 Permutations and Combinations
Last updated at May 29, 2023 by Teachoo
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 You are here
Example 12 (ii)
Example 13
Example 14 Important
Example 15
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 6 Class 11 Permutations and Combinations
Serial order wise
- Ex 6.1
- Ex 6.2
- Ex 6.3
- Ex 6.4
-
Examples
- Miscellaneous
Transcript
Example 12 Find the value of n such that nP5 = 42 nP3, n > 4 Given nP5 = 42 nP3 Calculating nP5 nP5 = 𝑛!/(𝑛 − 5)! = (𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)(𝑛 − 4)(𝑛 − 5)!)/(𝑛 − 5)! = n(n – 1)(n – 2)(n – 3)(n – 4) Calculating 42nP3 42nP3 = 42𝑛!/(𝑛 − 3)! = 42𝑛(𝑛 −1)(𝑛 − 2)(𝑛 − 3)!/(𝑛 − 3)! = 42n(n – 1)(n – 2) Now, nP5 = 42 nP3 n(n – 1)(n – 2)(n – 3)(n – 4) = 42n(n – 1)(n – 2) (𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)(𝑛 − 4) )/(𝑛(𝑛 − 1)(𝑛 − 2) ) = 42 (n – 3)(n – 4) = 42 n(n – 4) – 3(n – 4) = 42 n2 – 4n – 3n + 12 = 42 n2 – 7n + 12 = 42 n2 – 7n + 12 – 42 = 0 n2 – 10n + 3n – 30 = 0 n(n – 10) + 3(n – 10) = 0 (n – 10) (n + 3) = 0 So, n = 10, and n = – 3 n(n – 10) + 3(n – 10) = 0 (n – 10) (n + 3) = 0 So, n = 10, and n = –3 But, It is given in question n > 4 So n = –3 not possible Therefore, n = 10 only
