In how many ways can the following 11 letters: $A,B,C,D,E,F,X,X,X,Y,Y$ be arranged in a row so that every $Y$ lies between two $X$ (not necessarily adjacent)?
The "not necessarily adjacent" thing it getting everything messed up to me. I dont know how do I solve this. Any hint will be helpful.
Divide your letters into two sets, the specials $S=\{X,X,X,Y,Y\}$ and the rest $R=\{A,B,C,D,E,F\}$.
Because we're not allowing the $Y$s to be outside the $X$s, there are $3$ arrangements for the specials : $XYYXX, XYXYX, XXYYX$
The rest have $6!=720$ arrangements.
Finally we need to weave the two sets back together. We can put them together in $\binom {5+6}5 = 462$ ways (eg. like $rrssrssrrrs$ for $r\in R, s\in S$)
This gives $3\cdot720\cdot462 = 997920$ options.
Approach via cases and multiplication principle:
Given a specific $k$, continue with the following via multiplication principle:
Pick where the left-most $X$ is.
Pick location of the inner-most $X$.
Pick location of the $Y$'s
Pick what letter appears in every remaining unused location.
We get as a result $\sum\limits_{k=4}^{10} (11-k)(k-1)\binom{k-2}{2}6!=997920$, agreeing with the given books answer.
Shorter solution which matches the books given answer format:
Pick the locations occupied by $A,B,C,D,E,F$ This can be accomplished in $\binom{11}{6}\binom{11}{5}$ ways.
Pick which letter specifically occupies each of those chosen spaces. This can be accomplished in $6!$ ways
Now, the leftmost and right-most unused spaces must be occupied by $X$'s. Pick among the three remaining unused spaces which is occupied by an $X$. This can be done in $3$ ways.
This gives the answer of $\binom{11}{5}6!\cdot 3$
First step:
Don't distinguish between $X$ and $Y$. So how many different ways are there to to arrange $A,B,C,D,E,F,Z,Z,Z,Z,Z$? The answer is $$\frac{11!}{1!\cdot 1!\cdot 1!\cdot 1!\cdot 1!\cdot 1!\cdot 5!}=6\cdot 7 \cdots 11$$
Second step:
For each of such arrangements there is always the same number of ways to replace the $Z$ by $X$ and $Y$ so that the constraints are satisfied. This does not depend how the $A,B,C,D,E,F,Z$ are placed in the string. The most left $Z$ and the most right $Z$ cant be replaced by an $Y$ so must replaced by $X$. The remaining three inner $Z$ must be replaced by $X,Y,Y$. These are $$\frac{3!}{1!\cdot 2!}=3$$ possibilities.
So the result is
$$(6\cdot 7 \cdots 11)\cdot 3$$
AS OF NOW, THIS GIVES THE WRONG RESULT
The way I would do it would be to create the basic structure:
$$XYYX$$
and then count the number of ways to distribute the other letters over that, given that any of the remaining letters can be anywhere in that expression. To the left of everything, between $XY$, $YY$, $YX$, or to the right of everything. You must also be careful with counting distributions that give the same configuration. For example, adding an $X$ to the left and then an $X$ to the right gives the same as adding an $X$ to the right and then an $X$ to the left.
There are $5\cdot6\cdot7\cdot8\cdot9\cdot10$ ways of distributing all other $6$ letters, except for the third $X$. For example, $FAXBCYYDXE$ is such an arrangement. For the third $X$ how many places are there? Note that while you technically have $11$ spots, placing the third $X$ immediately to the left of immediately to the right of an $X$ bears no difference. To make it explicit, I will index the $Xs$: you have $FAX_1BCYYDX_2E$ and want to place $X_3$. Doing $FAX_1X_3BCYYDX_2E$ or $FAX_3X_1BCYYDX_2E$ is the exact same thing as far as letters are concerned. Thus there are $9$ different places. This makes $5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot9$ arrangements.