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Q.
The greatest common divisor ${ }^{31} C_3,{ }^{31} C_5$, $\ldots,{ }^{31} C_{29}$ is
Permutations and Combinations
Solution:
We have, for $3 \leq r \leq 29$
$r !(31-r) !\left({ }^{31} C_r\right)=31 !$
As the prime 31 divides R.H.S. and 31 does not divide $r$ ! and $(31-r)$ ! for $3 \leq r \leq 29$, 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 .