For any integer P greater than 1, P! denotes the product of all the in

I don't like the question, because it's not clear if order should matter. If we write 9! as (1)(9!), is that different or the same as writing it as (9!)(1)?

In any case, we can prime factorize 9!

9! = (9)(8)(7)(6)(5)(4)(3)(2) = (3^2)(2^3)(7)(2*3)(5)(2^2)(3)(2) = 2^7 * 3^4 * 5^1 * 7^1

Once you have a prime factorization, you can count all of a number's divisors by adding one to each exponent, and multiplying the resulting numbers. Here our exponents are 7, 4, 1 and 1. If we add one to each of these, we get 8, 5, 2 and 2, and multiplying these numbers, we get 160. With 160 different divisors, we can make 80 pairs of divisors that will multiply together to 9!, assuming order does not matter.

\(9!\)
=\(9 * 8 * 7 * 6 * 5 *4 * 3 * 2\)
=\(3^2 * 2^3 * 7 * 2 * 3 * 5 * 2^2 * 3 * 2\)
=\(3^4 * 3^7 * 7 * 5\)

So the number of factors of 9! = (4+1)(7+1)(1+1)(1+1)
= 5 * 8 * 2 * 2
= 160
So the number of ways by which 9! can be expressed as a product of 2 integer = 160/2 = 80. (D)

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Given: For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive.
Asked: In how many ways can 9! be expressed as a product of 2 positive integers?

\(9! = 1*2*3*4*5*6*7*8*9 = 2^7*3^4*5*7\)

Number of factors of 9! = (7+1)(4+1)(1+1)(1+1) = 8*5*2*2 = 160
Number of ways can 9! be expressed as a product of 2 positive integers = 160/2 = 80

IMO D