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  • If n is an integer between 0 and 21 then the minimum value of n and (21- n) is attained for n equals to

If n is an integer between 0 and 21 then the minimum value of n! and (21- n)! is attained for n equals to

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Multiply and divide n!(21-n)! by 21!, you get n!(21-n)! = \frac{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

Posted by

himanshu.meshram

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