Happy Easter! - 2

I tried to write this problem, and I've realized I've completely gone off the tracks.

If we randomly scramble the letters of HAPPYEASTER, in how many ways can exactly 2 sets of the duplicated letters appear next to each other in the string?

Example of desired outcome:

AAPYHSPEERT

#Combinatorics

Easy Math Editor

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Comments

Well, here's my brute-force solution. :) Running the following code (which could take around a minute due to the $\dfrac{11!}{2!2!2!} = 4,989,600$ initial permutations) yields a total number of ways of $\boxed{423,360}$ which is $\dfrac{14}{165}$ of the original permutations (for as yet unknown reasons).

# we'll need this for later
from itertools import permutations as perms

# create a list of all the letters in "Happy Easter"
letters = [l for l in 'HAPPYEASTER']
# create a list with itertools.permutations (imported as perms) of all possible permutations of the letters
# we also convert it to a set to remove duplicate permutations (since AA for instance will be counted twice)
letter_perms = set([p for p in perms(letters)])

# start counting permutations which meet our criteria
valid_perms = 0
# loop through every permutation
for perm in letter_perms:
    # start counting the duplicate sets (AA, EE, or PP)
    duplicates = 0
    # enumerate will give us the index of each letter in the permutation we're currently checking
    for i, letter in enumerate(perm[:-1]):
        # increases the duplicates count if we find a letter which is the same as the one after it
        if perm[i] == perm[i+1]:
            duplicates += 1
    # only increase the valid permutations count if there are exactly 2 duplicate sets
    if duplicates == 2:
        valid_perms += 1

# and this gives us our result!
print(valid_perms)

David Stiff - 2 months ago

Or, for those who prefer one liners... :)

1	`print([exec('from itertools import permutations as perms'), len([p for p in set([p for p in perms([l for l in 'HAPPYEASTER'])]) if len([p[i] for i,_ in enumerate(p[:-1]) if p[i]==p[i+1]])==2])][1])`

David Stiff - 2 months ago

Your code is correct!

Start by selecting 2 of 3 sets of duplicated letters to group.

That can be done in $\displaystyle {3 \choose 2}$ ways.

Next, the number of ways of arranging the string $H A_2 P P Y E_2 S T R$ is given by:

$\displaystyle {9 \choose 2} \cdot 7!$

We must take away the number of instances where the result above yields all three duplicated letter sets grouped by noticing the string $H A_2 P_2 Y E_2 S T R$ can be arranged in $8!$ ways:

This results in:

$\displaystyle {3 \choose 2} \cdot \left( {9 \choose 2} \cdot 7! - 8! \right) = 423,360$

Thank you @David Vreken for showing me the tactic in another post, and thank you for confirming by computational methods @David Stiff !

Eric Roberts - 2 months ago

So clever! I could have saved myself some pain by grouping the duplicate sets like that. It was that last step that I couldn't figure out. Thanks!

David Stiff - 2 months ago

Would it be possible for you to annotate so I can try to understand the code? I've never written in Python.

Eric Roberts - 2 months ago

Sure thing! I've updated the code above.

David Stiff - 2 months ago

I suppose it’s $(8+7+6+5+4+3+2+1)\times 3=108$ ways :) (possible positions times ways to choose 2/3)

Jeff Giff - 2 months ago

Really? Seems too simple. :)

I almost got it, but got stuck at the end. I figured there are $3$ different pairs of duplicate sets we can use, $72$ (I think) ways to arrange them, but then I couldn't figure out how many ways there are to arrange the remaining letters so that the last duplicate set didn't appear together...

Perhaps I should just code it up and find out. :)

David Stiff - 2 months ago

Ohh yes! Thanks for reminding me that I had to fill in the other letters as well :P For the rest 7 letters, there are $\frac{7!}{2}$ combinations! So the desired number is $216\times \frac{7!}{2}$ .
But we have to subtract the possibility of 3 pairs aligned :)

Jeff Giff - 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$