# Find number of integral solutions of $abcd=210$

Find number of integral solutions of $a\times b\times c\times d=210$

$$210=2\times 3\times 5\times 7$$ I tried by assuming 2,3,5,7 as numbered balls. The above problem is equivalent to placing 4 balls on 4 boxes where emplty boxes are allowed or placing 3 partitions between 4 balls. (Empty box signifies 1).

Assuming the partitions as sticks, I have to find the number of ways of arranging 4 different balls and 3 sticks. (The numbered balls between the sticks are like numbered balls in a box. So if two sticks come together, it means you get an empty box).

Number of ways = $7!$. But answer given is $8\times 4^4$

(I don't know if negative solutions are allowed. If that is the case, my method will not work. But if only positive integral solutions are allowed, is my method correct?)

Assuming only positive integral solutions, you’re assigning each of the $4$ primes to one of the $4$ ‘boxes’ $a,b,c$, and $d$. Both the primes and the ‘boxes’ are individually identifiable, so this can be done in $4^4$ ways. Thus, there are $4^4$ solutions in positive integers. However, the problem merely requires the four factors to be integers. We can assign plus and minus signs arbitrarily to $a,b$, and $c$, but then there will be only one possible choice of sign for $d$ in order to make the product positive, so there are altogether $2^3=8$ ways to assign the signs. Alternatively, an even number of $a,b,c$, and $d$ must be negative, and this can happen in $8$ ways: all positive, all negative, or one of the $\binom42=6$ ways of picking two to be negative.

Note that the problem is not equivalent to the usual one of placing $3$ partitions in a line of $4$ balls, because these ‘balls’ are individually identifiable. You can line them up in the order $2,3,5,7$ and place your three partitions in $\binom73$ ways, but that will only give you the factorizations in which no prime appears to the right of any larger prime. (E.g., you can’t get $7\cdot10\cdot3\cdot1$ this way.) Unfortunately, if you multiply by $4!$ to allow for all possible orders of the primes, you overcount: $1\cdot1\cdot1\cdot210$, for instance, gets counted $4!$ times!)

• I told I am taking permutations of 7 items (4 balls and 3 partitions). At one point, I will get: { 7 | 2 5 | 3 | } which is equivalent to your expression. – Aditya Dev Dec 30 '15 at 0:29
• Oh. In my method, there will be duplication. eg: {7 | 2 5 | 3 |} and {7 | 5 2 | 3 |} will be counted. so I end up with more than whats required. Is that correct? – Aditya Dev Dec 30 '15 at 0:34
• @Aditya: The three partitions are indistinguishable, so $7!$ is already too large by a factor of $6$. Unfortunately, even $\frac{7!}{3!}$ is quite a bit too large, since it counts $7\cdot10\cdot3\cdot1$ twice, once for $7\mid 2\,5\mid 3\mid$ and once for $7\mid 5\,2\mid 3\mid$. Without the division by $3!$ you’re counting that factorization $12$ times. \\ Yes, your new comment catches one of the two sources of overcounting. – Brian M. Scott Dec 30 '15 at 0:35