Discussion

Explanation:

(a + b) is greater than both a and b, hence (a + b)! will always be divisible by a! or b!.

Now, let us check a! × b!

$\frac{(a+b)!}{a!\times b!}$ = $\frac{1\times 2\times 3\times \dots \times a\times (a+1)\times (a+2)\times \dots \times b}{a!\times b!}$ = $\frac{(a+1)\times (a+2)\times \dots \times b}{b!}$

Now, (a + 1) × (a + 2) × … × b is product of b natural numbers, which will always be divisible by b!

∴ $\frac{(a+1)\times (a+2)\times \dots \times b}{b!}$ is an integer

⇒ $\frac{(a+b)!}{a!\times b!}$ is an integer i.e., (a + b)! is divisible by a! × b!

Hence, option (d).

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