In how many ways can be the letters of the word ELEEMOSYNARY be arranged so that the S is always immediately followed by a Y?

Attempt: There are 3 Es, and 2 Ys, and and then all letters appear once including the S. There are 12 letters in total Then assume SY is a letter, then there will be 10 letters. Then the word can ELEEMOSYNARY can be arranged in 10!/{3!7(1!)}. Please can someone please help me. I keep getting the wrong answer. Thank you.

  • $\begingroup$ You wouldn't multiply by $7$. The division is necessary because of the repetition of letters. If there is only one letter in the word, it wouldn't affect the number of arrangements. $\endgroup$ – Varun Iyer Sep 29 '14 at 23:43

Take SY to be one letter. You are now dealing with 11 letters instead of 12. So you have $$\frac{11!}{3!}=6,652,800$$ distinct arrangements.

If you look closely at the answer in the book, $$11\times\frac{10!}{3!}$$ is the same as $$\frac{11!}{3!}$$I think the book might have said, well, let's take away the SY as one letter. Now we are dealing with 10 letters we wish to permute. This gives $$\frac{10!}{3!}$$ And then there are 11 ways to slot the letter SY back into all the possible arrangements, so $$11\times\frac{10!}{3!}$$

  • $\begingroup$ Yes, but I still don't understand what that 11(10!) means? $\endgroup$ – user3459 Sep 29 '14 at 23:59
  • $\begingroup$ I made some edits. Does it make sense now? $\endgroup$ – Old mate Sep 30 '14 at 0:01
  • $\begingroup$ When you have all the possible arrangements of 10 letters, there are nine gaps in between each letter, and one gap on either end of the word where you can slot in the SY. So 9+1+1 total slots. $\endgroup$ – Old mate Sep 30 '14 at 0:03
  • $\begingroup$ Yes, thank you. $\endgroup$ – user3459 Sep 30 '14 at 0:04
  • 1
    $\begingroup$ No worries at all! Enumeration still does my head in! It's a tricky concept to think about sometimes. All the best. $\endgroup$ – Old mate Sep 30 '14 at 0:06

Here's how to attempt this problem.

Assume that the S and the Y are always together, meaning they form "one" letter.

Now we have that there are $11$ letters. Also, note that there is only technically $1$ Y, since the "letter" SY is included now. Now, we have that:

$$\frac{11!}{3!1!} $$

  • $\begingroup$ The answer of the book is 11[10!/3!] = 6,652,800. I don't understand why $\endgroup$ – user3459 Sep 29 '14 at 23:44
  • $\begingroup$ @user2349 I made some errors. After we group the S and Y together, the "total" number of Y's in the word is $1$, and therefore the overcount would result from the $3$ E's. Also, since the S and Y are together, we have only $11$ now. $\endgroup$ – Varun Iyer Sep 29 '14 at 23:47
  • $\begingroup$ That is another way, with the same 6652800 answer. But can someone please explain the book's answer? $\endgroup$ – user3459 Sep 29 '14 at 23:53

there are 12 letters so they can be arranged into $12$ factorial ways. However the letter decomposition of $eleemosynary$ is $a^1e^3l^1m^1n^1o^1r^1s^1y^2$ so each of the $12!$ arrangements fits into a group of $3!2!$ arrangements that look exactly the same. Thus there are $\dfrac{12!}{3!2!}=11!=39,916,800$ words

However as you said what we want is to turn $sy$ into a word so we get the word decomposition is $a^1e^3l^1m^1n^1o^1r^1y^1(SY)^1$ so there are $11!$ arrangements and each of them fits into a group of $3!$ arrangements that look exactly the same, thus we have $\dfrac{11!}{3!}=6,652,800$

  • $\begingroup$ sorry I'm late to the game, but my internet crashed and I had already written it $\endgroup$ – Jorge Fernández Hidalgo Sep 30 '14 at 0:04

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