First slide

Combinations

Question

The greatest common divisor $^{31} C_{3},^{31} C_{5}$ $\dots,^{31} C_{29}$ is

Moderate

A

31
B

2
C

17
D

none of these

Solution

We have, for $3 \leq r \leq 29$
$r! (31 - r)! (^{31} C_{r}) = 31!$
As the prime $31$ divides $R . H . S$ . and $31$ does not divide $r!$ and
$(31 - r)! for β \leq r \leq 29, we get 31 ∣ (^{31} C_{r})$
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 $31$ .

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