Correct option: (c) 52C4
Explanation:
47C4 + 5∑j=1 52–jC3
= 47C4 + 51C3 + 50C3 + 49C3 + 48C3 + 47C3 (1)
using nCr + nCr–1 = n+1Cr we get 47C4 + 47C3 = 48C4
∴ from (1) expression is 51C3 + 50C3 + 49C3 + 48C3 + 48C4
= 51C3 + 50C3 + 49C3 + 49C4
= 51C3 + 50C3 + 50C4
= 51C3 + 51C4
= 52C4