Math, asked by rathanrock, 9 months ago

find the minimum value of n!(21-n)!​

Answers

Answered by Manulal857
2

Answer:

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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