if n is an integer between 0 and 21 ,then the minimum value ofn!(21-n)! is _________________

Dear Aatish Agarwal,

Multiply and divide n!(21-n)! by 21!, you get n!(21-n)! = 21!/21 C n

so,minimum value of the given term is same as maximum of 21 C n

we know the max of N c r is for r = n-1/2 if n is odd

therefore for n= 10 or 11 {both same value}, 21 C n is maximum ,hence given term is minimum

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PLZZZ APPROVE MY ANSWER IF YOU LIKE IT

since minimum value of n!(21-n)! is the maximum value of 21Cn (=21!/n!(21-n)!), therefore it is equal to 10!(21-10)! or 11!(21-11)!, i.e. 11!10!.