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For example, how come $4 \choose 3$ (from 4 dice, choose 3 to be the same) can relate to the list:

d, s, s, s
s, d, s, s
s, s, d, s
s, s, s, d

Where s = same number, and d = different number? Shouldn't this be a permutations problem?

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    Because you chose exactly 3 out of 4: you chose 3 spots out of the 4 for the same number $s$.2012-03-29
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    If you are making four-letter "words" where you have only two available letters, indeed you are counting permutations. But the "choose" symbol can be useful in counting for example the number of words that have three s and one d. There is no one universal tool for counting permutations.2012-03-29
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    Actually binomial coefficient don't relate to permutations (even in the high-school sense of that word) but to combinations, which is exactly what you have here.2012-03-29

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