How to do counting problems with restrictions

Question

I have this homework question that I'm having trouble with:

In how many different orders can 10 runners finish a race if 3 people tie for the first place, 2 people tie for second place, 2 people tie for the third place, and no ties otherwise?

My first intuition is to try:

$ 10!/(10-3)! * 7!/(7-2)!*5!/(5-2)!*3! =720*42*20*6=3,628,800 $

But that is just equal to 10! when I feel that the answer should be less than 6! since there are only 6 possible places that the race can end in and there are restrictions on top of that.

Intuitively how are you supposed to do this type of a problem?

Edit: Updated possible ranks

Answer 1

Fortunately binomial coefficients capture the requirements: written $\binom mn $, say "$m$ choose $n$" and with value $\binom mn = \frac{\large m!}{\large n!(m-n)!}$.

So here we need $3$ in first place - $\binom {10}3 = 120$ - then $2$ in second place - $\binom 72 = 21$ - then $3$ in third place - $\binom 52 = 10$ - then the remaining $3$ runners chooes the final places - $3!$.

So the answer there is $\binom {10}3\binom 72\binom 52 \cdot 3! = 120\cdot21\cdot 10\cdot 6 = 151200$

You can see from the binomial coefficient formula that the difference from what you are trying is that we are also dividing to remove any order between the tied first-place finishers, for example.

Answer 2

Since you have $7$ runners tied for various places(from rank $1$ to $3$), only $3$ runners are left to have a rank of their own so you have a total of $6$(not $7$) distinct ranks - from $1$ through $6$.

Intuitively, you can choose any $3$ runners to share the first rank. Having done that you have $7$ runners left to choose from for the $2$ that share the second rank; then only $5$ left to pick $2$ from for the third and so on...

So you have the answer: $\binom{10}{3} \binom{7}{2}...$and so on till the sixth rank.

Answer 3

Let us first consider how many possibilities we have to compose first 3 ranks given we have chosen the $7$ runners. This $P=\frac{10!}{3!\cdot 2!\cdot 2!}$.

Denote $D=\binom{10}{7}$ the number of possible groupings for the first 3 ranks. Finally, $R=3!$ is the number of possibilities of arranging the remaining runners once the first 3 ranks are fixed.

The total is hence $R\cdot D\cdot P=151200$.