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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.

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30 = 2 * 3 *5 24= 2 * 2 *2 * 3 If that helps ?
1 is also there.
thats what i did too. what next?

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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 ?
Similar approach for 24

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