I’m not sure if I am making a mistake or if the key is wrong but I’m working a problem on a mathematics competition practice test which states:
How many distinct 5 letter arrangements can be made from ‘FLOYDADA’ with repetition allowed?
I’m assuming that the repetition allowed means that the D and A can be used twice and not that any letter can be used multiple times. To solve I separated into 3 cases, one where there are no repeated letters. One where there is one repeated letter, and one where 2 are repeated. Here is my solution:
No repeats: $6C5 \cdot 5! = 720$
One repeat: $2C1 \cdot 5C4 \cdot \frac{5!}{2!} = 600$
Two repeats: $2C2 \cdot 4C3 \cdot \frac{5!}{2!2!} = 120$
Altogether, I get $1440$ but the key says the answer is $2040$. Am I doing something wrong? I would really appreciate some help. Thanks!
Best Answer
You correctly counted the case in which there are no repeats. However, in the other two cases, you did not account for the number of positions that remain to be filled correctly after you selected the repeated letter(s). It looks like you obtained the right answer for the case with two repeats for the wrong reason.
Exactly one letter appears twice: There are two ways to select the letter that appears twice from among A and D, $\binom{5}{2}$ ways to select two positions for that letter, $\binom{5}{3}$ ways to select three of the remaining five letters, and $3!$ ways to arrange them in the remaining three positions. Hence, there are $$\binom{2}{1}\binom{5}{2}\binom{5}{3}3! = 1200$$ such arrangements.
Two letters each appear twice: The letters that appear twice must be A and D. Choose two of the five positions for the As, two of the remaining three positions for the Ds, and then choose which of the other four letters will fill the remaining position. Hence, there are $$\binom{5}{2}\binom{3}{2}\binom{4}{1} = 120$$ such arrangements.
Since there are $$\binom{6}{5}5! = 720$$ arrangements in which no letter is repeated, the number of distinct five-letter arrangements that can be made from the letters of FLOYDADA if a letter may be used as many times as it appears in FLOYDADA is $$\binom{6}{5}5! + \binom{2}{1}\binom{5}{2}\binom{5}{3}3! + \binom{5}{2}\binom{3}{2}\binom{4}{1} = 720 + 1200 + 120 = 2040$$