Question

i have 2 questions of similar type. looking for a permutation and combination method to it
Number of integral solutions of xyz=30 and xyz=24?
also explain what if it was positive integral solutions?
a start would be good.

Answer 1

30 = 2 * 3 *5
24= 2 * 2 *2 * 3

If that helps ?

Answer 2

1 is also there.

Answer 3

thats what i did too. what next?

Answer 4

factors of 30 = 1,2,3,5,6,10,15,30

Lets fix x=1 and say x<=y<=z
now (y,z) can be
(1,30)
(2,15)
(3,10)
(5,6)
No more.

If we fix x=2, (x<=y<=z)
(y,z) =
(3,5)
No more
If we fix x=3, (x<=y<=z)
(y,z) =None

So we come to a halt

Now note that x,y,z are all symmetrical, all x,y,z can be permuted in 3! ways except a special case when 2 are equal, then no. of permutations will be 3!/2!

Ans = (4*3!) + (1* 3!/2!) =27
Unless I'd have done something wrong ?

Answer 5

Similar approach for 24