# Throw 4 equal dice such that all of the faces show different numbers

Imagine one throws 4 equal dice at the same time. How many combinations with different faces (in which no face appears twice) are possible? For the first die one has 6 possible faces, for the second 5, for the third 4 and for the fourth 3.

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## 2 Answers

You have it if you think order matters. You multiply the number of choices at each stage. If you think 1234 is the same as 3142, you need to divide by the number of orders, here $4 \times 3 \times 2 \times 1=24$

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Ross Millikan got the logic right, but did the arithmetic wrong. (6*5*4*3)/(1*2*3*4) = 15, not 24.

Another way to look at it is to see 6C4 choices for the four numbers that appear, and give them 4! permutations. 6C4 * 4! = 6*5*4*3. (forgive the lousy notion for binomial coefficients; some day I will lurn to do it right.)

This problem is of interest to me because it is sort , of the complement of the problem I posted. For example, suppose you roll the die ten times; in how many of the 6^10 outcomes do all six numbers appear? 10C6 * 6^4 is not correct because it counts 1234561111 five times.

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Please read Ross Millikan's answer more carefully. He did not say that the answer was 24. He said that to correct the OP's answer one must divide by 24. – Pete L. Clark Jul 4 '14 at 22:09