# Permutation with Repetition when r<n

Question: In a game of chance after 12 rounds I have 3 wins and 9 losses. What are the possible number of sequence of outcomes after the first 5 rounds.

Approach: W W W and L L L L L L L L L So total ways of arranging this in 5 rounds is 12P5 and since there is repetition divide it by 3!x9!. Obviously there is something wrong with the approach. Could someone point that out and explain

Thanks

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Outcomes is sequence of wins and losses –  user88197 Jul 29 '13 at 19:25

Five Ls $1$ way; four Ls and one W $5$ ways; three Ls and two Ws $10$ ways; two Ls and three Ws $10$ ways. So $1+5+10+10 = 26$ ways in total. You could write this as

$$\sum_{l=\max(0,5-3)}^{\min(5,9)} {5 \choose l}.$$

Note that the different ways are not equally probable.

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Thanks Henry! Usually I use Total no. of arrangements without repetition / n1!xn2!..nk! (were n1,n2...nk is the # of repetitions). Could you tell if this is incorrect ! –  user88197 Jul 29 '13 at 19:40
@user88197 - that might work if you were arranging all the items, but here you are only arranging the first few. –  Henry Jul 29 '13 at 20:18