Three fair 6-sided dice are rolled.
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19 Aug 2021, 06:32
This question include 3 possibilities:
1. All 3 dice have same number
2. Two of the dice have same number
3. All the numbers are different
First let's calculate the combinations for 1st case. Since there are only 6 numbers we would have 6 cases (i.e. all 1s, 2s....6s).
Second, we calculate for two dice having same number such as two dice has 1s and then third die could have any number from 2-6. So 1*5 for this case. Therefore we would have 6*5 = 30 combinations of outcomes.
Third, let's consider the cases which would have all different numbers. This would be 6*5*4=120 but we need to eliminate permutation such as {1,2,3},{1,3,2},{2,1,3},{2,3,1},{3,2,1},{3,1,2} as order doesn't matter. There are 3!=6 ways that each groups of the numbers can be ordered so we need to divide 120 by 6 in order to obtain the results where order doesn't matter. Thus there are 20 combinations where all three numbers are different.
The total number of combinations are 6+30+20=56
Hence the answer is option C