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ABCDEABCDEABCDE is one (Posted on 2023-11-14) Difficulty: 3 of 5
How many arrangements of the letters AAABBBCCCDDDEEE are there such that there are no triple letters in the sequence?

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution solution | Comment 1 of 3
There are 13!/(3!^4) ways of AAA occurring.
Same for BBB, CCC, DDD, EEE.   Totaling 5*13!/(3!^4) = 24024000

There are 11!/(3!^3) ways of AAA and BBB.
Same for all C(5,2)= 10 double triples. Total is 10*11!/(3!^3) = 1848000

There are 9!/(3!^2) ways of AAA and BBB and CCC
Same for all C(5,3)= 10 triple triples. Total is 10*9!/(3!^2) =  100800

There are 7!/(3!) ways of AAA and BBB and CCC and DDD
Same for all C(5,4)= 5 quad triples. Total is 5*7!/(3!) = 4200

There are 5! ways of getting all of AAA, BBB, CCC, DDD and EEE. = 120


By inclusion/exclusion, ways of getting at least 1 triple = 24024000 - 1848000 + 100800 - 4200 + 120 = 22,272,720

Since we want no triples, the number of ways of that is 15!/3!^5 - 22272720 =  145,895,280.

Check  against a simulation:

In a million trials we'd expect around 132443 cases of finding at least one triple letter sequence. That way we can test the result with a simulation:

a='aaabbbcccdddeee';
ct=0;
for i=1:1000000
  b=a(randperm(15));
  found=false;
  for j=1:3:13
    f=strfind(b,a(j:j+2));
    if ~isempty(f)
      found=true;
    end
  end
  ct=ct+found;
end
fprintf('%d vs 132443',ct)

Three runs show

133132 vs 132443
132654 vs 132443
131935 vs 132443

a pretty good match.

A 10-million-trial run shows 1323536 vs 1324433.

  Posted by Charlie on 2023-11-14 14:46:20
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