Michelle Middle level student Question: Professional basketball, hockey, and baseball championships are decided on the best 4 out of 7 games. The first team to win four games wins the championship. In how many possible ways could the team win the championship series after winning the first game? ```Example: Game 1 2 3 4 5 6 7 Result W L L W W W ``` Hi Michelle, This question deals with what are known as "permutations" which are different orderings of a group of objects. In this case our "objects" of interest are "wins" and "losses". We know there are at most 7 games with at most 4 wins. This means there are 3 losses to make up the seven games. From the standpoint of what happens in real life, if 4 games in a row are won from the start, no more games are played, but for counting purposes, we would call those games not played losses. Now we need to look at how many ways we can order 4 wins and 3 losses. We shall denote a win as "W" and a loss as "L". You can try making a chart that lists them all: LLLWWWW LLWLWWW LLWWLWW and so on. But this is really not a practical way to approach it because there will be many possible orderings and it is easy to get confused and list the same ones twice or even miss a possible ordering. Thankfully in math there is a formula for counting in these types of situations (permutations with repeated symbols). In general, if there are n objects of one type, and m objects of another type, and n + m = p, then there are  p!/n!m! arrangements of the p objects. In case you are not familiar with the factorial symbol (!), n! = (n)(n-1)(n-2)...(2)(1) In our question we have p = 7, n = 4 and m = 3 and thus we have orderings or arrangements. This does not, however, answer your specific question because you have been told that the first game was a win. The 35 arrangements we counted include all of those that also start with a loss. Can you see how to modify what I showed you here to solve your specific problem? Leeanne Go to Math Central