My Math Forum Combinations where only one element can repeat.

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 February 25th, 2017, 09:37 PM #1 Newbie   Joined: Feb 2017 From: Brazil Posts: 1 Thanks: 0 Combinations where only one element can repeat. Hi! Let's say I have the letters A, B, C, D. I want to know the number of combinations in the space _, _, _, or 3, where repetitions are allowed. For this, I can apply the formula $\displaystyle C=n ^ x$. In the example would be $\displaystyle C=4 ^ 3=64$. My question is: How do I calculate this, but if only one letter can be repeated? Let's just say "A" can be repeated. For instance: A, A, A A, B, A C, A, A D, A, B (This one does not repeat any, but is part of the result) ... ... Thanks!
 February 25th, 2017, 10:03 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,753 Thanks: 896 Does order matter? does $ABD=DAB$ ? If not... this will be the sum of the number of combinations with no repeats plus the number of combinations with 1 repeat. There are $\binom{4}{3}=4 \text{ combinations with 0 repeats.}$ There are $\binom{4}{1}\binom{3}{1}= 12 \text{ combinations with 1 repeat.}$ $\text{Thus a total of 4+12=16 combinations allowing a single repeat.}$
February 25th, 2017, 11:53 PM   #3
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Quote:
 Originally Posted by romsek Does order matter? does $ABD=DAB$ ? If not... this will be the sum of the number of combinations with no repeats plus the number of combinations with 1 repeat. There are $\binom{4}{3}=4 \text{ combinations with 0 repeats.}$ There are $\binom{4}{1}\binom{3}{1}= 12 \text{ combinations with 1 repeat.}$ $\text{Thus a total of 4+12=16 combinations allowing a single repeat.}$
hrm from your formula I see that order does matter.

In this case all we have to do is just subtract off those combos where elements are repeated 3 times.

There will be 4 of these.

So there will be $4^3 - 4 = 60$ different combinations allowing a single repeat.

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