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Find the number of different arrangements that can be formed using all the letters of the word PAPAYA. Among these arrangements how many contain the three letters ‘A’s in consecutive order?

I have tried 6!/2!3!

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  • $\begingroup$ The number $\frac{6!}{2!3!}$ counts the number of distinguishable arrangements of the letters of the word PAPAYA. Hint: Think of AAA as a single letter. $\endgroup$ – N. F. Taussig Oct 23 '17 at 16:06
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think of it as having six slots. You need to find all the ways you can stick P's A's and a Y into those slots.

Youve got 3 A's. So there are $\binom{6}{3}$ ways you can put the A's in. After that, youve got 3 slots left, and two P's. So youve got $\binom{3}{2}$ ways to put the P's in. And then finally you have $\binom{1}{1}$ way to arrange the final Y. It doesnt matter the order you do these in, you could have started with the Y, and then the P and then the A's and instead of $\binom{6}{3}\binom{3}{2}\binom{1}{1}$ youd have $\binom{6}{1}\binom{5}{2}\binom{3}{3}$ and it ends up being the same answer, which is 60.

So of those 60 possible arrangements, have many have AAA? Well if we think of "AAA" as a single letter, now we have 4 slots to put the letters "AAA", "P", "P", and "Y" into, and no matter how we arrange them "AAA" will be together. So $\binom{4}{1} \binom{3}{2} \binom{1}{1}$

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    $\begingroup$ Be careful. In the second problem, you have not taken into account the fact that two of the four slots are filled with Ps. $\endgroup$ – N. F. Taussig Oct 23 '17 at 16:20
  • $\begingroup$ @N.F.Taussig oops, good point! Edited to fix, thanks $\endgroup$ – XRBtoTheMOON Oct 23 '17 at 16:26
  • $\begingroup$ Your answer is still incorrect. I suggest that you approach the problem the way you did the first one. You have four objects to arrange. Choose two of them for the Ps, choose one of the remaining two positions for the AAA, and then place the Y in the remaining slot. $\endgroup$ – N. F. Taussig Oct 23 '17 at 16:38

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