Expert: Josh - 3/12/2008

Question
QUESTION: Based your level of Math expertise this should be very easy for you.  Actually I do believe it is easy for me but since the answer key says we are wrong, I would like someone else to solve it as well.

I am hoping you will read this math problem and tell us what you come up with and how, I won't give you our answer so not to influence the outcome.
What we come up (which I think is correct) is not what the answer key says.


 Jeff, Lorri, and Dan had a race. In how many different ways could they come in first, second, and third?

We appreciate you telling us your answer and HOW you came up with your answer.

ANSWER: Hi Mom

We can treat this as a combinatorial problem. For this question, however, I think it is easier to just list all possible outcomes.

Key Assumption: We cannot allow two people tying for 1st place, or 2nd place. Also, we cannot have three winners.

Writing the names of 1st, 2nd and 3rd place winner in order from left to right, we get
(J,L,D)
(J,D,L)
(L,J,D)
(L,D,J)
(D,J,L)
(D,L,J)
So, there are six distinct possibilities.

If we treat this as a combinatorial problem, we can ask "how many ways can we pick the winner"? Ans: 3 (from J, L or D)

Case 1: If J got first place, we are left with L and D. We can then pick either L, or D as the runner-up. And the matter of who gets third place is automatically decided. This leads to 2 possibilities.

Case 2: If L got first place, we are left with J and D. We can then pick either J or D as the runner-up. Again, whoever we select for the runner up, it is a certainty that the other person will be in third place.

Case 3: If D got first place, we are left with J and L. We can then pick either J or L as the runner-up. Again, whoever is left behind will be in the third spot.

The situation in all three cases are symmetrical. Ultimately, the 1st, 2nd and 3rd positions can be determined in 3x2x1 = 6 ways.

Can I watch TV now? :)

---------- FOLLOW-UP ----------

QUESTION:  Now I have a problem, I get 9, 3 x 3.  I assume that any of the three could come in first, second or third, so it would be 3 x 3.  Even though you gave a detailed explanation, I don't get it. :(

Answer
You are thinking that you are picking 1 out of 3 people for the first place, then, picking 1 out of another 3 people for the second place. This is not the situation depicted in the question.

The flaw is that a person cannot hold two or more places at the same time. e.g., if J got first place, that leaves L and D fighting for the second place. J is out of the running. So, you don't have {J,L,D} to choose from for the second place. Instead, you can only put either L or D in second place. This can happen in two ways.

Look at it this way. How many ways can we pick a winner from {J,L,D}?
Ans: 3 ways. Because the winner must be J, L or D.

What happens next? Whoever got first place is out of the running for the second place. It doesn't matter who got first place, the situation is the same. Let's just say that L got first place for the sake of argument. This leaves J and D competing for 2nd place. How many ways can we pick the runner up from {J and D}?
Ans: 2 ways.

What happens next? The outcome is fixed. If J had finished 2nd, then it is a certainty that D finished 3rd; and vice versa. Assuming we disallow a tie for 2nd place.

So, we have three ways of selecting a winner, and two ways of selecting the runner-up. Multiplying these together, we get 3x2 = 6 ways of arranging J, L and D in 1st, 2nd and 3rd place.

The possible outcomes are
1st: J. 2nd: L. 3rd: D
1st: J. 2nd: D. 3rd: L
1st: L. 2nd: J. 3rd: D
1st: L. 2nd: D. 3rd: J
1st: D. 2nd: J. 3rd: L
1st: D. 2nd: L. 3rd: J

The only thing I can suggest is don't pull numbers out of a hat, or whatever comes to mind. Try reasoning the process of selection step by step as we have done above. The truth will come out:)

Hey, talk to someone about this. May be they can explain it to you in another way.