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Factors of 12 are : 1 , 2, 3, 4, 6 which are single digits. so a, b, c are from 1,2,3,4,6.

Combinations/groups of digits which result in 12 : 1 , 2 , 6, ;; 2, 2, 3 ;; 1 , 3, 4 ;;;

For the 1st and third groups of these 3 digit groups, there are 3! or 3P3 permutations possible. So there are 6 numbers in each group. Hence, there are 12 such numbers. In the second group, as two digits are same, the permutations are 3! /2 = 3 only.

hence there are totally 15 such numbers.

these are : 126, 162, 216,261,612,621,;; 223,232;;322 ;;;; 134, 143, 314, 341, 413, 431

Product of three no.can be 3 ✖ 4 ✖ 1
There is 3-choices for a,2 choices for b and and 1 choice for c
choices for results is 3 multiply ✖ 2 ✖ 1 equals to 6
We can also break a,b,c as 3 ✖ 2 ✖ 2
These have same procedure as mention above, then total type it formed is 6
We can also break as 6,2,1
Apply same procedure
We will also get same as above 6
Add total choices 6+6+6=