# A permutation/ race problem

• November 6th 2012, 10:51 AM
zzizi
A permutation/ race problem

There are 12 children in a race wearing different coloured hats. 4 Red, 4 blue and 4 white hats.

How many possible outcomes of the race are there in which 3 of the first 5 children that cross the line are wearing red hats?

I would really appreciate an explanation too. Thank you!
• November 6th 2012, 11:36 AM
HallsofIvy
Re: A permutation/ race problem
Start by looking just at the first five racers. Imagine that the first three are wearing red hats. There are four possibilities for the first child, then 3 for the second, two for the third. There are 4(3)(2)= 24 ways that can happen. Now, there are 8-non-red hats so 8(7)= 56 ways the fourth and fifth students can be "non-red hats". That would give a total of (56)(24)= 1344 ways the first 5 children can finish "red, red, red, non-red, non-red". But there are also 5!/(3!2!)= 10 different orders for "RRRNN" so, in fact, 13440 ways the first five children could be "3 red hat, 2 non-red hat". Now, we have 7 children remaining. There are 7! ways they can finish.

(This is assuming that we are talking about the ways individual children, not just colors of hats, can finish. That is, "Jimmy first, Sarah second" is different from "Sarah first, Jimmy second" even though both are wearing red hats.)
• November 6th 2012, 12:24 PM
Plato
Re: A permutation/ race problem
Quote:

Originally Posted by zzizi
There are 12 children in a race wearing different coloured hats. 4 Red, 4 blue and 4 white hats.
How many possible outcomes of the race are there in which 3 of the first 5 children that cross the line are wearing red hats?

I assume that you are to count the number of different ways the children can finish the race in which three red hats are among the first five places.
$\left[\binom{4}{3}\binom{8}{2}(5!)\right]\left[7!\right]$
We select the three red hats and two non-red hats and then arrange the first five; then arrange the other seven.
• November 15th 2012, 11:01 AM
zzizi
Re: A permutation/ race problem
Thank you very much for your responses.

Quote:

$\left[\binom{4}{3}\binom{8}{2}(5!)\right]\left[7!\right]$
Can you just verify, what is the calculation for this? Are you multiplying these together?

I received another answer from someone else I posed the question to and it differs from the answers given here, so am alittle confused.

He stated the following:

if we number the children as 1 2 3 4 || 5 6 7 8 || 9 10 11 12
we want combos in which 3 out of 5 are from the 1st group
possible color combos are RRRBB RRRWW & RRRBW
considering distinct #s, combos*perms for RRRBB = 4c3*4c2*5!/(3!*2!) = 240
ditto for RRRWW -> 240
for RRRBW -> 4c3*4c1*4c1*5!/3! = 1280

total possible outcomes = 2*240 + 1280 = 1760
• November 15th 2012, 11:32 AM
Plato
Re: A permutation/ race problem
Quote:

Originally Posted by zzizi
Can you just verify, what is the calculation for this? Are you multiplying these together?

Here is the OP "How many possible outcomes of the race are there in which 3 of the first 5 children that cross the line are wearing red hats?"

Now that is poorly posed.
I took it quite literally. The order of the children is all important here. Say Bob has a red hat and Jim has a red hat. If Bob is first and Jim is third is totally different from Jim first and Bob second. Although in both cases those two are among the first five to finish.

That is the basis of my reply. If the order of the finishers makes no difference then we must ask. Do you simply mean how many ways can the first five be wearing three red hats and two non-red hats? And that is regardless of order of any kind.

Or what exactly does the question mean?
• November 15th 2012, 11:44 AM
zzizi
Re: A permutation/ race problem
Quote:

Originally Posted by Plato
Here is the OP "How many possible outcomes of the race are there in which 3 of the first 5 children that cross the line are wearing red hats?"

Now that is poorly posed.
I took it quite literally. The order of the children is all important here. Say Bob has a red hat and Jim has a red hat. If Bob is first and Jim is third is totally different from Jim first and Bob second. Although in both cases those two are among the first five to finish.

That is the basis of my reply. If the order of the finishers makes no difference then we must ask. Do you simply mean how many ways can the first five be wearing three red hats and two non-red hats? And that is regardless of order of any kind.

Or what exactly does the question mean?

Apologies if it was poorly posed, that's how it was asked of me. That's why I put the question here, so that I could make better sense of what is meant and how to solve. Thank you for your input though, it's appreciated. :)
• November 15th 2012, 12:17 PM
Plato
Re: A permutation/ race problem
Quote:

Originally Posted by zzizi
Apologies if it was poorly posed, that's how it was asked of me. That's why I put the question here, so that I could make better sense of what is meant

If we assume that only the order of the colors is all that matters then the answer is
$\frac{5!}{3!\cdot 2!}\cdot\frac{7!}{(4!)(3!)}=350$
• November 15th 2012, 12:19 PM
zzizi
Re: A permutation/ race problem
Quote:

Originally Posted by Plato
If we assume that only the order of the colors is all that matters then the answer is
$\frac{5!}{3!\cdot 2!}\cdot\frac{7!}{(4!)(3!)}=350$

That's great thank you
• November 17th 2012, 11:52 AM
zzizi
Re: A permutation/ race problem
Quote:

Originally Posted by Plato
If we assume that only the order of the colors is all that matters then the answer is
$\frac{5!}{3!\cdot 2!}\cdot\frac{7!}{(4!)(3!)}=350$

I'd just like to ask a clarifying question:

when you are multiplying by: $\ffrac{8!}{(4!)^2}=700$
Can you explain to me how you derived the 4!^2?
• November 17th 2012, 12:15 PM
Plato
Re: A permutation/ race problem
Quote:

Originally Posted by zzizi
I'd just like to ask a clarifying question:
when you are multiplying by: $\ffrac{8!}{(4!)^2}=700$
Can you explain to me how you derived the 4!^2?