a, b, and c are integers from 1 to 4, inclusive. What is the probability that a≤b≤c?

Suppose you fix a = 1. What options do you have?

1 2 3

1 3 4

1 2 4

these are the possible combinations if we fix a=1.

2 3 4 if we fix a=2.

total combinations are 4 * 3 * 2=24

4/24 should be the probability right?

1 2 2 is also an option. It isn’t stated that the numbers have to be distinct.

**This may not be the best way to solve this question.**

I think you need to brute force every combination here. I can’t think of a better way.

So by that method,

If you take a = 1, you can have

bc as 11, 12, 13, 14, 22, 23, 24, 33, 34, 44 (10 choices)

Now for a = 2, you can not have 11, 12, 13, 14 (10 - 4 = 6 choices)

For a = 3, you can not have the above 4 and 22, 23, 24 ( 6 - 3 = 3 choices)

Finally, for a=4 you can only have 1 choice which is 44

Allowed choices = 10 + 6 + 3 + 1 = 20

Total numbers = 4x4x4 = 64

So probability is 20/64 = 5/16