Question

The greatest common divisor ${}^{31}{C}_3,{}^{31}{C}_5$, $\dots ,{}^{31}{C}_{29}$ is

• A. 31
• B. 2
• C. 17
• D. none of these

Answer 1

Answer: (A)

We have, for $3\le r\le 29$
$r!(31-r)!\left({}^{31}{C}_r\right)=31!$
As the prime 31 divides R.H.S. and 31 does not divide $r$ ! and $(31-r)$ ! for $3\le r\le 29$, we get $31\mid \left({}^{31}{C}_r\right)$.
Also, since ${}^{31}{C}_{29}=(31)(3)(5),{}^{31}{C}_3=(31)(29)(5)$ and ${}^{31}{C}_5$ $=(31)(29)(7)$
No prime other than 31 can divide all the numbers.
Thus greatest common divisors of the given numbers is 31 .