Number of solutions

If $a,b,c,d,e \ge 2$ then the left side is at most $(3/2)^5$ which is less than $14.4$ . So at least one of the variables is $1$ . Suppose $e = 1$ for simplicity.

Then a similar argument using $(3/2)^4 < 7.2$ shows that at least one of the remaining four variables is $1$ , and a similar argument using $(3/2)^3 < 3.6$ shows that one of the remaining three is $1$ .

So we are left with $\left( 1+\frac1a \right) \left( 1 + \frac1b \right) = 1.8$ . Neither of the variables can be $1$ because the left side would be too big, but if they are both at least $3$ then the left side is at most $(4/3)^2 < 1.8$ . So one of them is $2$ . Then we're left with $1+\frac1a = 1.2$ and so $a=5$ .

This shows that the unique solution up to permutations is $(5,2,1,1,1)$ . There are $\fbox{20}$ permutations of this solution.

Multiply by $5abcde$ to get $5(a+1)(b+1)(c+1)(d+1)(e+1)=72abcde$ 5 is a prime number, so at least one of the factors on the right hand side must be an integral multiple of 5. So,without loss of generality we can set $a=5k$ , where $k$ is a positive integer.

Also, set $b\ge c,d,e$ . We get $(5k+1)(b+1)(c+1)(d+1)(e+1)=72kbcde$

Because $2c \ge(c+1)$ , etc., we can replace $(c+1)$ on the left hand side with $2c$ and replace the equal sign with $\ge$ . Doing this for $d$ and $e$ too, we get

$(5k+1)(b+1)8cde \ge 72kbcde$ which can be rewritten as $b \le \frac{5k+1}{4k-1}$

If $k=1$ we get $b\le2$ and if $k > 1$ we get $b <2$ .

We had chosen $b \ge c,d,e$ , so the choice for k limits these variables as well.

Plugging $b=c=d=e=1$ into the first equation, we'd get $80(a+1)=72a$ , so this gives no integral solution.

The only other option is $k=1, b=2$ , in which case we just find the tuple $(5,2,1,1,1)$ and all its $\boxed{20}$ permutations