Pemutations and combinations

Question

In case, find n:

a) nP4 = 30 nC4.

b) nCn−2 = 10.

Answer 1

nCn-2 = 10

nCr = n!/r!(n-r)!

nCn-2 = n!/(n-2)!(2!) = 10

n(n-1)/2 = 10

n^2 -n = 20

n^2 -n-20 = 0

(n-5)(n+4) = 0

n = 5

5C3 = 5!/3!2! = 5(4)/(2) = 10

nP4 = 30

nPr = n!/r!

nP4 = n!/4! = 30

n! = 30(4!) = 30x4x3x2 = 720

n!= 720

6! = 720

n = 6

nC4 = n!/4!(n-4)! = n(n-1)(n-2)(n-3)/4(3)(2)= (n^3-3n^2+2n)(n-3)/24

or

nC4 =nP4/2! = 30/2 = 15 = 6C4, with n=6