If (n+1)C(n-1) = 28, find the value of n

Question

C stands for Combinations

Answer 1

(n+1)C(n-1) = 28
(n+1)! / ((n+1-n+1)! (n-1)! ) = 28
(n+1)! / (2! (n-1)! ) = 28
(n+1)(n)(n-1)!/(2! (n-1)! ) = 28
n(n+1)/2 = 28
n^2 + n = 56
n^2 + n- 56 = 0
(n-8)(n+7) = 0
n = 8 or n = -7 , but n ≥ 0

n = 8

another way would be to realize that n+1 and n-1 have a difference of 2.
So in your Pascal's Triangle go down in the third column to find 28
1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 ....

look at the number in front of 28

Answer 2

Thanks!!

Answer 3

but n^2 + n - 56 = (n+8)(n-7)

Answer 4

Thanks Steve, good catch.