Afair die is rolled four times. find the probability that all four rolls show different numbers

Number of such outcomes, in which each number is no less than the preceding number, will be equal to number of ways of selecting 4 numbers from 6, with replacement; without considering order!

Why?

Lets choose any such set of size 4, say {2, 1, 1, 3}. we can sort it to get a sequence {1,1,2,3} which is one of our desired outcomes. So each such sorted sequence corresponds to one selection of size 4, with replacement and without considering order.

Number of such selections will be equal to number of solutions of following equation:

x1 + x2 + x3 + x4 + x5 + x6 = 4

where:

x1: number of 1 in the selections

x2: number of 2s in the selection

.

.

x6: number of 6s in the selection

Number unique solutions of such equation = 9 choose 5 = 126

(For more details on this see : Unordered sampling with replacement )

Number of possible outcomes = 6^4 = 1296

Probability of favorable outcomes = 126/1296=7/72

umber of possible outcomes = 6^4 = 1296

Probability of favorable outcomes = 126/1296=7/72

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step-by-step explanation:

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