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#1

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#2

@kunalsaini8950 hey approach could be like this, factor n! into p and q such that hcf(p,q)=1 and p<q. so we calculate number of distinct primes in n!
let prime factorisation of n!= (p1)^x1 * (p2)^x2 …(pk)^xk
now for each i, (pi)^xi can either go in p or q. hence 2^k ways. but as we want p<q. therefore only 2^(k-1) ways.

#5

@kunalsaini8950 hey here is code snippet for this problem:
https://ideone.com/ukVA8T

#7

@kunalsaini8950 bro we are not allowed to share our number ap chat pr puch lo.

#8

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