shankar.ashwin wrote:How many different combinations of outcomes can you make by rolling three standard (6-sided) dice if the order of the dice does not matter?
"¨"¨(A) 24
"¨(B) 30
"¨(C) 56
"¨(D)120
"¨(E) 216
I like the approach that cans took, but it took me a moment to determine the logic she used for 2 numbers are the same. So, I thought I'd flesh out the solution a bit more and offer another approach.
case 1: three different numbers
There are 6 possible values and we want to choose 3 of them. Since order doesn't matter, this is a combination question.
We can choose 3 numbers in 6C3 ways (=
20)
case 2: two numbers are the same
Cans' approach:
Step 1: first choose any 2 numbers. This can be accomplished in 6C2 ways.
Step 2: At this point we have 2 different numbers, so to get 2 numbers the same, we need to duplicate one of those numbers. There are 2 numbers to choose from, so this can be accomplished in 2 ways.
So, the total number of ways to accomplish steps 1 and 2 is 6C2 x 2 (=
30)
Another approach:
Step 1: select the number that you wish to have 2 of. There are 6 numbers to choose from, so this step can be accomplished in 6 ways.
Step 2: select the number that you wish to have 1 of. At this point, there are 5 numbers remaining, so this step can be accomplished in 5 ways.
So, the total number of ways to accomplish steps 1 and 2 is 6 x 5 (=30)
case 3: all three numbers are the same
Choose which of the 6 numbers you wish to have 3 of.
This can be accomplished in
6 ways (or we can say it can be accomplished in 6C1 ways)
So, the total number of ways =
20 +
30 +
6 = 56
Cheers,
Brent
