mm007 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
When we roll 3 dice, we could have the following 3 cases:
1) 3 of the same numbers
2) 2 of the same numbers and 1 different number
3) all 3 different numbers
Let’s analyze each of these cases.
1) 3 of the same numbers
If all 3 numbers are the same, they can only be (1, 1, 1), (2, 2, 2), ... , (6, 6, 6). Thus, there are 6 outcomes in this case.
2) 2 of the same numbers and 1 different number
If 2 numbers are the same and the third one is different, they could be, for example, two 1s and one of the 5 other numbers. However, we can replace the two 1s with any of the 5 numbers. In other words, we have 6 choices for the 2 numbers that are the same and 5 choices for the number that is different. Thus, there are 6 x 5 = 30 outcomes in this case.
3) all 3 different numbers
Since we are picking 3 numbers from 6 in which order doesn’t matter, there are 6C3 = 6!/[3!(6-3)!] = (6 x 5 x 4)/3! = (6 x 5 x 4)/6 = 20 outcomes in this case.
Therefore, there are 6 + 30 + 20 = 56 outcomes in total.
Answer: C
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