+680 In how many ways can 36 be written as the product \(a \times b \times c \times d\) , where \(a, b, c\) and \(d\) are positive integers such that \(a \leq b \leq c \leq d\)?
+1005 do it yourself!
List all the combinations then arrange then to the format!
Guest gave a wonderful hint!
+310 After solving the problem, I'm not really sure how this hint ties into the problem.
Are you saying there is 1 option for a, 2 options for b, 3 options for c, and 6 options for b? Or did you factorize 36?
Just curious to see if your solution was maybe faster than casework.
+680 I'll think about complementary counting, haven't tried that method on this problem yet...
Thanks for the idea
+310 There aren't that many cases because of the \(a \le b \le c \le d\) restriction, so we don't have to worry about permutations.
Using casework we find:
Case 1: 3 ones (36 as 1 factor)
1*1*1*36
Case 2: 2 ones (36 as 2 factors)
1*1*2*18
1*1*3*12
1*1*4*9
1*1*6*6
Case 3: 1 one (36 as 3 factors)
1*2*3*6
1*2*2*9
1*3*3*4
Case 4: 0 ones (36 as 4 factors)
2*2*3*3
Counting them up, we find that there are 9.
