Henry Ernest Dudeney/Puzzles and Curious Problems/276 - An Effervescent Puzzle/Solution

by : $276$

 * An Effervescent Puzzle

Solution
The answer to the first question is: $88 \, 200$ ways.

The answer to the second question is: $6300$ ways.

Proof
First we ignore the $\text E$'s and arrange the rest.

Consider the remaining letters as:
 * $\text F_1 \text F_2 \text {R V S}_1 \text {C S}_2$

There are $7!$ ways to arrange these letters.

Since the $\text F$'s and $\text S$'s are supposed to be indistingishable, we should divide the number of ways they can swap around to undo double-counting.

Therefore there are:
 * $\dfrac {7!}{2! 2!}$

ways to arrange these letters.

Now we insert the $\text E$'s inbetween these letters, or before and after these letters.

There are $8$ available spaces to put these $4$ $\text E$'s in.

There are, therefore:
 * $\dfrac {7!}{2! 2!} \times \dbinom 8 4 = 88 \, 200$

ways to arrange these letters in a line where no $\text E$'s are directly next to each other.

For the second question, we first arrange the letters, without $\text E$'s, on a single line.

As above, there are:
 * $\dfrac {7!}{2! 2!}$

ways to arrange these letters.

Now we insert the $\text E$'s inbetween these letters.

However, unlike when we did this on a line, the two ends, when wrapped in a circle, are actually the same space, so two $\text E$'s cannot occupy them at the same time.

Out the the $\dbinom 8 4$ possibilities, $\dbinom 6 2$ of them would be in this undesired arrangement.

(We cannot use $\dbinom 7 4$, i.e. consider the ends as the same space, as they are distinct spaces when we consider shifting.)

There are, therefore:
 * $\dfrac {7!}{2! 2!} \times \paren {\dbinom 8 4 - \dbinom 6 2} = 69 \, 300$

ways to arrange these letters in a line where no $\text E$'s are directly next to each other.

Now we wrap this line around in a circle.

In a circle, shifting all letters over by one space would not make a difference.

There are $11$ ways to shift the letters for each arrangement (including not shifting).

So, when wrapped around in a circle, there are:
 * $\dfrac {69 \, 300} {11} = 6300$

ways to arrange these letters where no $\text E$'s are directly next to each other.