find the minimum value of n!(21-n)!
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Hey Buddy here's ur answer.
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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