# How can I determine the amount of combinations where no spot can filled by more than one object? (b)

a) An NFL Committee has 38 members. How many ways are there to choose three members to serve on the subcommittee?

b) the NFL executive has four positions: commissioner, vice-commissioner, secretary and treasurer. How many ways are there to choose a commissioner, vice-commissioner, secretary and treasurer where no person can hold more than one office?

c) Roger is an executive of the NFL. How many solutions in (b) have Roger as the commissioner?

I've got $C(38,3) = 38!/3!(35!) = 8436$ ways to choose the NFL subcommittee and for (b) I have $C(38,4)-C(37,3)-C(36,2)-C(35,1) = 65380$ ways to choose where no person can hold more than one office. But this number seems like to many to me? If anyone could provide a good explanation of how to tackle this problem I would be greatly appreciative.

• If the committee had six members, could you answer part b) just by creating a decision tree of all the possibilities or by listing them all? Sep 6, 2018 at 22:24

I think part b) is somewhat easier to solve than part a) intuitively, so I'll start with b)

If we were to think of all of the possibilities, we can first, think about how many ways we can pick the president, then commissioner, vice-commissioner, and finally treasurer.

So, there are $38$ people we can pick for the president, then since there are no people holding two positions, there are $37$ for commissioner, $36$ for vice-commissioner, and $35$ for treasurer. So we take their product (P.S. This can be more easily written as a $\textbf{permutation}$, as $_{38}P_4$) to get $\color{red}{1771560}$.

Gigantic, huh? Let's take part a) now. We could take the same approach, and calculate $_{38}P_3$ and get $50616$. However, we've seriously overcounted. In this case, if I pick John first and Amy second, that's the same as picking Amy first and John second. So, we have to divide our results by the number of ways of rearranging our three choices, i.e., $3!=3\cdot2\cdot1$. This is known as a $\textbf{combination}$, and is often expressed as $_{38}C_3$ or $38\choose3$.

So, the answer to part a) is $\color{red}{8436}$.

For part c), we return to our work for part b). We counted every way to make the committee. Now how many of those cases have Robert as commissioner? Well, we could have picked anyone of $38$ guys, who are all identical, so we simply can divide our answer from part b) by $38$ to get our answer of $\color{red}{46620}$.

Again, these numbers are gigantic because for each choice we make, we can make another nearly independent choice for another position. That means a lot of crossover cases, making any attempts to count these cases by hand futile.

• I understand this now. For some reason I was thinking that (b) meant that if one person was commissioner then they would not have a position in any other scenario. Thank you for the help Sep 6, 2018 at 23:01

Well, informally speaking, $38$ is a big enough number to create so many possibility for choosing out of $38$ items (or people). So yes, the answers here are quite big, and that's okay.

That being said, while your answer to (a) is perfectly correct, your answer to be (b) isn't. To be honest, I don't understand at all the logic behind that expression that you set up…

The key difference between questions (a) and (b) is that in (a) order doesn't matter, but in (b) it does. In (a) we just need to select three people, but there's no ordering on them — nobody is first or second or third. Selecting John and Jack and Joe is the same as selecting Joe and Jack and John is the same as … you get the idea. That's why finding the combinations $C(38,3)$ was exactly the right thing to do.

But in (b) order matters, there's the first person (commissioner), second person (vice-commissioner), third person (secretary), and fourth person (treasurer). Selecting John as commissioner and Jack as vice-commissioner and … is different from selecting Jack as commissioner and John as vice-commissioner and … So what you need here is permutations, not combinations. And the formula for the number of permutations of $r$ objects out $n$ when repetition is not allowed is $\displaystyle P(n,r)=\frac{n!}{(n-r)!}$.

In this example, you can get the answer using this formula, or we can apply some basic logic. Here's a similar example. Given a club of $5$ people, in how many ways can we choose a president and a secretary for this club if they can't be the same person? Well, if we start with the president, then we have $5$ options (people) to choose from. Then to fill the secretary position we have only $4$ available people, because one is already unavailable (selected as president). So all in all, we have $5\cdot4=20$ possibilities.