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

Note by Eric Roberts
5 months, 1 week ago

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Well, here's my brute-force solution. :) Running the following code (which could take around a minute due to the 11!2!2!2!=4,989,600\dfrac{11!}{2!2!2!} = 4,989,600 initial permutations) yields a total number of ways of 423,360\boxed{423,360} which is 14165\dfrac{14}{165} of the original permutations (for as yet unknown reasons).

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# 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 - 5 months, 1 week ago

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

Eric Roberts - 5 months, 1 week ago

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Sure thing! I've updated the code above.

David Stiff - 5 months, 1 week ago

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Your code is correct!

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

That can be done in (32) \displaystyle {3 \choose 2} ways.

Next, the number of ways of arranging the string HA2PPYE2STR H A_2 P P Y E_2 S T R is given by:

(92)7! \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 HA2P2YE2STR H A_2 P_2 Y E_2 S T R can be arranged in 8! 8! ways:

This results in:

(32)((92)7!8!)=423,360 \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 - 5 months, 1 week ago

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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 - 5 months, 1 week ago

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

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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 - 5 months, 1 week ago

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

Jeff Giff - 5 months, 1 week ago

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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 7!2\frac{7!}{2} combinations! So the desired number is 216×7!2216\times \frac{7!}{2}.
But we have to subtract the possibility of 3 pairs aligned :)

Jeff Giff - 5 months, 1 week ago

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Really? Seems too simple. :)

I almost got it, but got stuck at the end. I figured there are 33 different pairs of duplicate sets we can use, 7272 (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 - 5 months, 1 week ago

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