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ISI2021-MMA: 12

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Answer : (C) 64

$\prod_{k=1}^{5} k! = 1! *2!*3!*4!*5!$
$ = 1*2*1*3*2*1*4*3*2*1*5*3*2*1 = 2^8 *3^3 *5$

$\text{Total number of divisors } = (8+1)*(3+1)*(1+1) = 72$

$\text{total number of odd divisors} = (3+1)*(1+1) = 8$

$\text{total number of even divisors} = \text{total number of divisors} – \text{total number of odd divisors}$

$72 – 8 = 64$

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