The number of three element sets of positive distinct integers {a, b, c} such that the product abc = 15015 is? Q. The number of three element sets of positive distinct integers {a, b, c} such that the product abc = 15015 is ?
My approach:-
15015 = 1 * 3 * 5 * 7 * 11 * 13
Now I thought of this problem as 6 distinct balls (1, 3, 5, 7, 11, 13) to be distributed in 3 distinct boxes (a, b, c), such that each box at least have 1 ball in them.
So, I made the following possibilities:-
Case 1:- 3  2  1  ---> $6C3 * 3C2 * 1C1 * 3!$ = 180
Case 2:- 2  2  2 ----->  $6C2 * 4C2 * 2C2 $  = 90
Case 3:- 4  1  1 -----> {6C4 * (2C1 * 2C1)/2!} * 3!} = 90
Giving a total of 360 cases, but the answer given is 40, please help me to identify mistake and how to proceed ?
 A: First consider ordered solutions.  There are $5$ distinct prime factors.  Allocate each one to a slot...that's $3^5$.  Now, it's fine if one slot is empty but we don't allow two empty slots (as the three numbers are required to be distinct).  Thus we must remove $3$ cases.
Thus we have $3^5-3=240$ ordered solutions.  As we can permute any solution in exactly $6$ ways, the answer must be $\frac {240}6=40$.
A: Ignore the number $1$ in your analysis.  We have the primes $3,5,7,11,13$
We have the possible breakdowns: $4$-$1$-$0$, $3$-$2$-$0$, $3$-$1$-$1$, and $2$-$2$-$1$.  Note that $5$-$0$-$0$ doesn't count since this would have resulted in a two element set, not a three.
For the case of $4$-$1$-$0$, pick which element goes for the second batch.  $5$ ways.
For the case of $3$-$2$-$0$, pick which three primes go for the first batch... $\binom{5}{3}$.  The remaining primes go for the second.
For the case of $3$-$1$-$1$, choose the three primes to go for the first batch.  The remaining primes go one each into the others and it doesn't matter which.  $\binom{5}{3}$ here again.
For the last case of $2$-$2$-$1$, choose the prime that goes for the $1$.  Now... among the remaining four primes one will be smallest and might as well be sent to the first batch.  Choose which of the remaining primes goes with that smallest prime.  $5\cdot 3$ ways.
$$5+\binom{5}{3}+\binom{5}{3}+5\cdot 3 = 5+10 + 10 + 15 = 40$$
