sachin_yadav wrote:How many positive integers less than 10,000 are such that the product of their digits is 30?

A. 12
B. 24
C. 36
D. 38
E. 50

OA is E

Please share your approach

First, let's lay out the ways we can multiply several positive digits (numbers between 1 and 9) to get 30. (Because the number must be less than 10,000, we know that we want fewer than five digits.)

6 * 5 * 1 will work --> There are 6 ways we can arrange these 3 digits. (Any of the 3 numbers can be in the hundreds place. Any of the 2 remaining numbers can be in the tens place. And the remaining number must be in the units place: 3*2*1 = 6)

3*2*1*5 will work --> There are 24 ways we can arrange these 4 digits: 4*3*2*1 = 24

6*5*1*1 will work ---> There are 12 ways we can arrange these 4 digits. (If all 4 digits are different, there are 24 arrangements. But swapping the 1's will not result in unique numbers, so we need to divide by two to account for the fact that the two 1's are interchangeable.)

6*5 will work ---> 2 ways to arrange these 2 digits

2*3*5 will work ---> 6 ways (3*2*1) to arrange there 3 digits

Now add up the numbers in red: 6 + 24 + 12 + 2 + 6 = 50

Casework (as David showed) is probably the best way; this is an approach that is very common for GMAT-level combinatorics questions.