If 12 (15C1 ) + 22 (15C2 ) + 32 (15C3 ) + + 152 (15C15 ) = 2m 3n 5k, where m, n, k N, then m + n + k is equal to :
- 20
- 19
- 18
- 21
Answer
(2) 19
Solution
For NCr: r=1N r2 NCr = r=1N [r(r - 1) + r] NCr
= N(N - 1)2N-2 + N 2N-1
= N(N + 1)2N-2
For N = 15: sum = 15 16 213
= 3 5 24 213 = 217 31 51
m = 17, n = 1, k = 1
m + n + k = 19
Answer: (2)