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.
On
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:
$$\frac{11!}{3!1!} $$
On
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 $\dfrac{11!}{3!}=6,652,800$
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 $$11\times\frac{10!}{3!}$$