Hi everyone,

I'm a bit confused on the best approach to the "Basketball Teams" problem on a recent MGMAT CAT. Here's the question:

"John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?"

I read through the approach, and while it makes sense, it isn't exactly the most intuitive solution (at least not to me). I know that the first part of the problem should be to do an anagram of YYYYYNNNN, but the second part of the solution (i.e. assume John and Peter are already on the team and come up with another anagram of 7 plays - YYYNNNN) is still unclear. Are there any other ways to tackle this problem?

Thanks,
Shuo

If you're not a fan of the anagram method, here is a more academic approach.

Probability is basically the number of possibilities we want divided by the number of total possibilities.


Let start with the number of total possibilities.

Number of total possibilities: Use Combinations.
Out of 9 players, we choose 5

9 choose 5
-> 9 C 5
--> 9! / (4! * 5!)

Note, you arrive at the same equation using your anagram YYYYYNNNN


Let's move on to the number of possibilities we want.

Number of possibilities we want: All possible teams that include John and Peter.
A team has 5 people, and 2 of them are John and Peter. Therefore, we need to find all possible combinations for the other 3 empty spots. Since John and Peter are already chosen, we have 9-2=7 possible players that can fill those 3 spots. Use Combinations again.

Out of 7 players, we choose 3

7 choose 3
-> 7 C 3
--> 7! / (3! * 4!)

Note, you arrive at the same equation using your anagram YYYNNNN


You can solve by using these two values. I hope this makes things more clear for you.

Thanks tmmyc!

i'm still a bit confused. i actually like the anagram method; it's just that i'm confused as to why you would have an anagram of YYYNNNN for the second portion of the problem. in other words, i'm unclear on why you would "assume" that the two spots have already been reserved for john and peter already and only consider the 7 other folks.

thanks guys.

Ah, got it. So, we assume that John and Peter are given "Y"s because the question specifically asks us about the probability that they will both be on the team. Essentially, we're just setting up the anagram to match the question: the question indicates that we want John and Peter both to be on the team, so we assign them each a Y. If they had asked us, instead, for the probability that neither would be on the team, we would assign them each an N. Does that make sense?

Here's a simple way to think of it:

In problems that order doesn't matter do the following

(Total # of people/things)! / (# of positions open)! x (Total # of people - # of positions open)!

First let's find the total # of teams of 5 people that can be selected from 9.

Using the equation from above we get 9! / 5! x 4! = 126

So now we know there are 126 teams to choose from and some of them have the two guys and some of them don't.

Since order doesn't matter a team with John, Peter, Nick, Mike, and Bob is no different from the team with Bob, Nick, Peter, Mike, and John. It's still the same team.

So right now we want to know how many different teams there are with the 2 guys on it.

We agreed above that the order of the players doesn't matter as long as John and Peter are on the team. So let's put them on the team and find out how many different way we can fill up the rest of the spots with the 7 players we have left.


John--Peter--(Open spot)---(Open spot)---(Open spot)


So remember our equation from above: (Total # of people/things)! / (# of positions open)! x (Total # of people - # of positions open)!

7! / (3! x 4!) = 35


35 out of the 126 different teams contain John and Peter.

Hope this helps!

Thanks, Spencer. Yes, you can use either the anagram method or the classic comb/perm method we learned back in high school - whatever works best for you!