Question

Number of cyphers at the end of ${}^{2002}{C}_{1001}$ is

• A. 0
• B. 1
• C. 2
• D. 200

Answer 1

Answer: (B)

1

Consider given the given expression${}^{2002}{C}_{1001}$.

For this we just need to find the highest power of 10 in numerator and denominator,

Numerator =$2002!$

Denominator =$={(1001!)}^2$

For numerator we will get $499249$ zeroes.

Therefore , ${(1001!)}^2$ will have $498$ zeroes.

Now, the solution just a cake walk

$\frac{499}{498}=1$

Hence, this is the answer.