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If a 6-sided die is rolled 5 times and each roll is recorded as an element of set A (|A| will be 5 after all rolls), How many results out of all the possible results will have exactly two 4's as elements.

First step was obviously to figure out how many possible results there are to begin with, Since the result will be a sequence of 5 items from 6 choices, with repeats allowed I used the formula (n+k-1)Ck, so here it will be 10C5, 10!/(5!5!) = 252 possible results.

I'm stumped how to figure out how many of these results contain exactly two 4's.......

I feel like the answer is staring me in the face but I can't quite see it lol. Just looking for a nudge in the right direction!

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Total possibilities is not 252 it is $6^5$ , 6 options for each roll. Secondly exactly 2 4's , chose any two positions $\binom{5}{2}$ rest each of 3 positions has 5 options so answer is $\binom{5}{2}*5^3$.

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  • $\begingroup$ woops, yeah I don't know why I thought to use the other formula here! $\endgroup$ – Mike Jan 31 '15 at 19:24
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    $\begingroup$ Yes I tried, I had to wait 4 more minutes....... $\endgroup$ – Mike Jan 31 '15 at 19:34
  • $\begingroup$ Interestingly, about 1/6th of the results contain exactly two 4's. haha. $\endgroup$ – Mike Jan 31 '15 at 19:59

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