A number N when expressed as product of prime factors gives.

Question is quite ambiguous but here is what I think:

So we are looking for all cases of N = x * y * z like:

N = (1) * (1) * (2^5 * 3^2 * 5^4 * 7 * 11^3);
N = (1) * (2^5) * (3^2 * 5^4 * 7 * 11^3);
N = (1) * (2^5 * 3^2) * (5^4 * 7 * 11^3);
N = (2^5) * (3^2) * (5^4 * 7 * 11^3);
N = (2^5 * 3^2) * (5^4 * 7) * (11^3);
...

In this case factors x * y * z would be co-prime (wont share any common factor but 1) as each prime will be only in one factor.

As there are total of 5 primes in N then there are following cases possible:

1. {1}*{1}*{factor with all five primes} - 1 (1*1*N);

2. {1}*{factor with one prime}*{factor with four primes} - \(C^1_5*C^4_4=5\) (\(C^1_5\) - choosing which prime will be in one-prime factor, the rest primes go to the third factor);

3. {1}*{factor with two prime}*{factor with three primes} - \(C^2_5*C^3_3=10\) (\(C^2_5\) - choosing which 2 primes will be in two-prime factor, the rest primes go to the third factor);

4. {factor with one prime}*{factor with one prime}*{factor with three primes} - \(C^3_5=10\) (\(C^3_5\) - choosing which 3 primes will be in three-prime factor, from the 2 primes left one will go to the first factor and another to the second);

5. {factor with one prime}*{factor with two primes}*{factor with two primes} - \(C^1_5*\frac{C^2_4}{2}=15\) (\(C^1_5\) - choosing which 1 primes will be in one-prime factor, 4 primes left can be split among two factors (into two groups) in \(\frac{C^2_4}{2}\) ways, dividing by 2 as the order of the factors (groups) does not matter);

Total: \(1+5+10+10+15=41\).

No such answer among answer choices.

So I guess it's meant that the order of the factors is important, for example N= (2^5) * (3^2) * (5^4 * 7 * 11^3) is different from N= (3^2) * (2^5) * (5^4 * 7 * 11^3), then case 1 can be arranged in 3 ways and cases 2, 3, 4, and 5 in 3! ways: \(3*1+3!(5+10+10+15)=243\).

So answer: C.

But if the order of the factors is important then the problem can be solved easier: each of the five primes can be part of the first, the second or the third factor so each prime has 3 options. Total ways to distribute 5 primes would be \(3*3*3*3*3=3^5=243\).

Answer: C.

Anyway not a GMAT question, so I wouldn't worry about it at all.