How different 6 letter "words" can you form using one a, two b's, and three c's?
I've approached this question two different ways and get different answers, one method must be wrong, but I can't seem to convince myself of which is wrong.
Method 1: There are 6! ways to permute the 6 letters, but since there are two indistinguishable b's and three indistinguishable c's, we have a total of $\frac{6!}{2!3!} = 60$ ways.
Method 2: Pretend there are 6 "slots" for each letter in the 6-letter word. There are $\binom{6}{1}$ ways to pick the slot for letter 'a'. Once 'a' has been put into a slot, there are $\binom{5}{1}$ ways to pick the slot for letter 'b' and another $\binom{4}{1}$ ways to pick the slot for the second 'b'. The remaining three slots are fixed and left for the letter 'c' to fill in. So there are $\binom{6}{1}\binom{5}{1}\binom{4}{1} = 120$ ways in total.
I have a feeling method 2 is wrong, but I can't seem to pinpoint where I've gone wrong conceptually.
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$\begingroup$Method two is wrong, because you have double counted the possible positions for $b$. For instance, putting an $a$ in slot 1, a $b$ in slot 2, and then a $b$ in slot 3 results in the same word as putting an $a$ in 1, then a $b$ in 3, and then a $b$ in 2.
You have counted each word precisely twice (namely, for each order of the $b$s), so you should divide the final answer by 2 to get the correct answer.
Another way of putting this is that instead of choosing one position twice, as $\binom{5}{1}\binom{4}{1}$, you instead choose two positions at once, as $\binom{5}{2}$. This also gets you the correct answer.
$\endgroup$ 1 $\begingroup$you can select 2 places from 6 and place a's on there which is $\binom{6}{1}$ and find places that b's can go by $\binom{6-1}{2}$ and c can go by $\binom{6-2-1}{3}$ which is $\binom{6}{1}$$\binom{5}{2}$$\binom{3}{3}$ . We can multiply them which will give us $ \frac{6!*5!*3!}{1!*5!*2!*3!*3!}$ it is equal to 60. We can also type it as a multinomal coefficent, it will be $\binom {6}{3,2,1}$ which is identical to first one
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