The number of letters between $z$ and $y$ is less than that between $y$ and $x$

Question

QUESTION: Consider all the permutations of the $26$ English alphabets that start with $z$. In how many of these permutations the number of letters between $z$ and $y$ is less than that between $y$ and $x$?

Source: ISI BMath 2017 UGA

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MY APPROACH: This is what I have done-

The $13^{th}$ place is the mid point of the $26$ places we have for the $26$ letters. Now, the first place is obviously occupied by $z$. Observe that,

If $y$ sits somewhere right to the $13^{th}$ place then $x$ cannot be placed to the right of $y$, and therefore, $x$ will only occupy any position to the left of $y$. So, that's not required.

Note From here on we will not mention that $x$ can always be placed at the left of $y$ (since that's unnecessary). We will only bother with the cases where $x$ is to the right of $y$.

If $y$ occupies the $13^{th}$ place then $x$ can occupy the $26^{th}$ place only, therefore we get only one way for $x$ here.

If $y$ occupies the $12^{th}$ place then $x$ can be placed at $26^{th}$,$25^{th}$ or $24^{th}$ place, thus $3$ ways.

Following this pattern, we get the series (as $y$ moves from $12^{th}$ to the $2^{nd}$ place) $$1+3+5+...+23$$

So, we are done with all possible cases where $x$ is to the right of $y$.

And therefore,, our answer must be $$[144].23!$$

Am I correct? If not, where is the mistake that I have made?

Can you please provide me with some alternate solution? This seems a bit lengthy..

Thank you so much.

Answer 1

That’s definitely bad — if x is to the left of y, the number of letters between x and y is certainly less than that between z and y, which is the wrong way around!

(this is because the sequence must start with z)

On the other hand, $144\times 23!$ looks right to me, if you disinclude those cases.

Answer 2

Think of 26 numbered place holder slots in which you can fit letters. In the first slot, I put the letter $z$, after this, I think as to where to put the letter $y$ so that my placement will be in accordance with the condition of question.

Clearly, I can not keep it any further than the 13th slot from $z$, then the following criteria in the problem statement will be failed (as also noted by OP):

The number of letters between $z$ and $y$ is less than that between $y$ and $x$

Now, let us count for each placement, how many possible placements of $x$ are possible.

If we keep $y$ in the second slot, then we have twenty three slots in which we can keep $x$. Twenty three and not twenty four because, again, the condition in question would be vioded if we kept the $x$ right after $y$

If we keep $y$ in the third slot, then by discounting the invalid slots of the remaining twenty four, we find that there would be twenty one placements for $x$ (remember the distance between y and x must be strictly greater)

If we keep $y$ in the fourth slot, we find there would be 19 slots for $x$

... ..

if we keep $y$ in the nth slot, we find there would be $26- \left[n+(n-2) +1\right]= 26 - 2 \left[ n-1 \right]$ slots where we can keep $x$

The total placements of $x$ require summing up all the possible cases till $n=32$ i.e: $$ \sum_{n=2}^{13} 26 - \left[2 n-1 \right]= 144$$

Now, the total way to arrange the twenty three other letters for all of these cases are 23!. Hence, we have a total of $ 144 \cdot 23!$ arrangements