27720 = 2³*3²*5*7*11
There are 5 prime factors. Since the two factors of 27720 must be coprime, if we
choose a 2 for one factor to contain, we must choose 2³ for that factor to
contain. Similarly if we choose a 3 for one factor we must choose 3² for that
factor to contain. So it depends on how many prime factors, not how many times
the prime factors are contained in 22720.
Case 1:
Number of ways we can choose 1 prime factor for the first and 4 prime factors
for the second.
That's 5C1 = 5 ways
1. first factor = 2³, second factor = 3²*5*7*11
2. first factor = 3², second factor = 2³*5*7*11
3. first factor = 5, second factor = 2³*3²*7*11
4. first factor = 7, second factor = 2³*3²*5*11
5. first factor = 11, second factor = 2³*3²*5*7
Case 2:
Number of ways we can choose 2 prime factors for the first and 3 prime factors
for the second.
That's 5C2 = 10 ways
1. first factor = 2³*3², second factor = 5*7*11
2. first factor = 2³*5, second factor = 3²*7*11
3. first factor = 2³*7, second factor = 3²*5*11
4. first factor = 2³*11, second factor = 3²*5*7
5. first factor = 3²*5, second factor = 2³*7*11
6. first factor = 3²*7, second factor = 2³*5*11
7. first factor = 3²*11, second factor = 2³*5*7
8. first factor = 5*7, second factor = 2³*3²*11
9. first factor = 5*11, second factor = 2³*3²*7
10. first factor = 7*11, second factor = 2³*3²*5
Answer 5C1 + 5C2 = 5 + 10 = 15 ways.
Edwin
