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2015-08-26T03:22:13+05:30

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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

1 5 1
2015-08-26T03:25:53+05:30
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=
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