Combinatorics Example Problem

For the second, you pick the slots (not the people-we said all the people from one country were interchangeable) for the two US people in the bottom ${3 \choose 2}$ ways, but then have to pick which slot the US person in the top is in, which adds a factor ${3 \choose 1}$. Then of the seven left, you just have $\frac {7!}{4!2!}$ as there are no Americans among them. So Method 1 is off by the $3!$ in the denominator. I don't understand the terms in Method 2.

Added: One way to look at the first part, which I think is the way you did, is to say there are $10!$ orders of the people, but if the Americans are interchangeable you divide by $3!$, for the Russians you divide by $4!$ and for the Chinese you divide by $2!$. Another way would be to say we pick three slots for Americans in ${10 \choose 3}$ ways, then pick slots for the Russians in ${7 \choose 4}$ ways, then pick slots for the Chinese in ${3 \choose 2}$ ways, giving ${10 \choose 3}{7 \choose 4}{3 \choose 2}=120\cdot 35 \cdot 3=12600$. Of course the answer is the same, but the approach is better suited to the second part. Now if you think of choosing slots for nationalities you get what I did before.

Part 2 is solved in two stages (using Method 1):

Stage 1: Let U represent a slot holding a US competitor, and let _ represent a slot holding a non-US competitor. The first three slots must be one of the following:

U _ _

_ U _

_ _ U

This is where the $\binom{3}{1}$ comes from, since we are choosing one of three slots to be the U slot. The middle four slots have to be _ _ _ _. There is only one way to do this—we could write this as $\binom{4}{0}$ since we are choosing zero of the four slots to be U slots, or we could simply omit this factor. The last three slots must be one of the following:

U U _

U _ U

_ U U

This is where the $\binom{3}{2}$ comes from: we are choosing two of three slots to be U slots.

Stage 2: We now have a slot arrangement, of which _ U _ _ _ _ _ U U _ is an example. (There are a total of nine possibilities.) We now have to fill the blank slots with the three non-US nationalities. There are $\frac{7!}{4!\,2!\,1!}$ ways to do this.

Note: Since we only care about which nationality occupies which slot, and not which individual competitor, at no stage should we be selecting individual competitors from the set of competitors of a given nationality. I think this is where you went wrong. (You selected two particular US competitors from the set of three US competitors. There is no reason to do this.)

Alternative solution: I think Method 1 is a bit confusing because in Stage 1 we are choosing slots, whereas in Stage 2 we are permuting nationalities and dividing out repeats. A more uniform method would be to think of the entire process as one of choosing slots. If we take this point of view, there there are

• $\binom{3}{1}$ ways to choose one of the first three slots for the US.
• $\binom{3}{2}$ ways to choose two of the last three slots for the US.
• $\binom{7}{2}$ ways to choose two of the slots not yet assigned for China.
• $\binom{5}{4}$ ways to choose four of the slots not yet assigned for Russia.
• $\binom{1}{1}$ way to choose the remaining unassigned slot for Canada.

I believe this is what's going on in Method 2, although they wrote the factors in an unnatural order.