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I have a question,

In how many ways can $6$ tosses of a coin yield $2$ heads and $4$ tails?

Now, to me, the question clearly seems to be of a permutation, as they have asked for the number of ways (they have mentioned nowhere that we have to choose.) But the actual solution tells that the question is solved by a combination.

Why is it a question of combination and not of permutation?

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    You are choosing, but in a somewhat subtle way: there are six tosses, which we can label 1, 2, 3, 4, 5, 6. You have to choose two of these to be heads, which can be done in ${}_6C_2$ ways, and you have to choose four of the four remaining tosses to be tails, which can be done in only ${}_4C_4=1$ way. Hence there are ${}_6C_2\cdot1$ ways.2012-06-15

6 Answers 6