Permutation with Repetition when r

Question

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

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

score 0 · Answer 1 · answered Jul 29 '13 at 19:30

0

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.

answered Jul 29 '13 at 19:30

Henry

157,058

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

Permutation with Repetition when r

1 Answers1