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GMAT/Problem Solving - Three Blind Dice
Three Blind Dice
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
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| Решение |
The three-dice combinations fall into 3 categories of outcomes:
1) All three dice have the same number
2) Two of the dice have the same number with the third having a different number than the pair
3) All three dice have different numbers
By calculating the number of combinations in each category, we can determine the total number of different possible outcomes by summing the number of possible outcomes in each category.
First, let’s calculate how many combinations can be made if all 3 dice have the same number. Since there are only 6 numbers, there are only 6 ways for all three dice to have the same number (i.e., all 1’s, or all 2’s, etc.).
Second, we can determine how many combinations can occur if only 2 of the dice have the same number. There are 6 different ways that 2 dice can be paired (i.e., two 1’s, or two 2’s or two 3’s, etc.). For each given pair of 2 dice, the third die can be one of the five other numbers. (For example if two of the dice are 1’s, then the third die must have one of the 5 other numbers: 2, 3, 4, 5, or 6.) Therefore there are 6 x 5 = 30 combinations of outcomes that involve 2 dice with the same number.
Third, determine how many combinations can occur if all three dice have different numbers. Think of choosing three of the 6 items (each of the numbers) to fill three "slots." For the first slot, we can choose from 6 possible items. For the second slot, we can choose from the 5 remaining items. For the third slot, we can choose from the 4 remaining items.
Hence, there are 6 x 5 x 4 = 120 ways to fill the three slots. However, we do not care about the order of the items, so permutations like {1,2,5}, {5, 2, 1}, {2, 5, 1}, {2, 1, 5}, {5, 1, 2}, and {1, 5, 2} are all considered to be the same result. There are 3! = 6 ways that each group of three numbers can be ordered, so we must divide 120 by 6 in order to obtain the number of combinations where order does not matter (every 1 combination has 6 equivalent permutations).
Thus there are  combinations where all three dice have different numbers.
The total number of combinations is the sum of those in each category or 6 + 30 + 20 = 56. The correct answer is C.
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John has more money than Mary but less than Bill. If the amounts held by John,
Mary, and Bill are x, y, and z, respectively, which of the following is true?
A) z < x < y
B) x < z < y
C) y < x< z
D) y < z < x
E) x < y < z
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Вверх 
