# 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?)

## 1 Answer

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