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There is a set of 19 characters A,B,C,. . . ,S. What is the maximum number of words of 16 characters that can be created from them if we require them to contain the character A is present once, character B is present twice, character C is present three times, and the other characters at most once? (regardless of whether they make sense).

My idea is to do permutations of a multiset: $\frac{16!}{1!*2!*3!}$ and the result is $120$. Is it right?

baranovak
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1 Answers1

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Choose the $10$ characters from the remaining $16$ ones and then permute (each choice is the same multiset "type" (i.e. the amounts are same):

$$ \binom{16}{10} \frac{16!}{3! \cdot 2! \cdot (1!)^{11}} = 13962475118592000. $$

ploosu2
  • 8,707
  • But we choose 10 characters from remaining 16, or not? – baranovak Dec 01 '23 at 21:09
  • @baranovak Oh yes, I got the numbers mixed up. – ploosu2 Dec 02 '23 at 08:11
  • And is it really right? I also tried this calculation: ${16 \choose 1}\cdot {15 \choose 2}\cdot {13 \choose 3}\cdot \frac{16!}{(16-10)!}$ – baranovak Dec 02 '23 at 12:56
  • Yes, that gives the same answer! So in your way you're like choosing the places for C, B and A and then the rest is a 10-permutation from 16 (i.e. choosing the remaining with order mattering and then putting them in that order to the free places not used by A's, B's and C's. – ploosu2 Dec 02 '23 at 13:22
  • Yes, that´s true! thank you so much! – baranovak Dec 02 '23 at 13:56