Consider the number `N=2016`.
Q. Number of cyphers at the end of `.^(N)C_(N//2)` is

Options

A. 0
B. 1
C. 2
D. 3

Answer

C

Explanation

`because .^(N)C_(N//2)=.^(2016)C_(1008)=((2016)!)/([(1008)!]^(2))`
`E_(5)(2016)!=[(2016)/(5)]+[2016/5^(2)]+[2016/5^(3)]+[2016/5^(4)]`
`=403+80+16+3=502`
and `E_(5)(1008)!=[1008/5]+[1008/5^(2)]+[1008/5^(3)]+[1008/5^(4)]`
`=201+40+8+1=250`
Hence, the number of cyphers at the end of `.^(2016)C_(1008)`
`=502-250-250=2`