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Twelve contestants at the county fair have

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by gmat_winter » Wed May 20, 2015 11:48 pm
Twelve contestants at the county fair have entered their cakes to be judged in the cake decorating competition. A purple ribbon, blue ribbon, red ribbon, and white ribbon will be given to the first, second, third, and fourth place competitors, respectively. How many different ways are there to award the four ribbons to the contestants?

A) 8! / 4!4!
B)8!/4!


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by theCEO » Thu May 21, 2015 12:47 am
gmat_winter wrote:Twelve contestants at the county fair have entered their cakes to be judged in the cake decorating competition. A purple ribbon, blue ribbon, red ribbon, and white ribbon will be given to the first, second, third, and fourth place competitors, respectively. How many different ways are there to award the four ribbons to the contestants?

A) 8! / 4!4!
B)8!/4!


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If possible please list the other choices. Often times problems can be solved by process of elimination.
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by GMATGuruNY » Thu May 21, 2015 2:06 am
gmat_winter wrote:Twelve contestants at the county fair have entered their cakes to be judged in the cake decorating competition. A purple ribbon, blue ribbon, red ribbon, and white ribbon will be given to the first, second, third, and fourth place competitors, respectively. How many different ways are there to award the four ribbons to the contestants?
Number of options for the purple ribbon = 12. (Any of the 12 contestants.)
Number of options for the blue ribbon = 11. (Any of the 11 remaining contestants.)
Number of options for the red ribbon = 10. (Any of the 10 remaining remaining contestants.)
Number of options for the white ribbon = 9. (Any of the 9 remaining contestants.)
To combine these options, we multiply:
12*11*10*9 = [spoiler]12!/8![/spoiler].
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by outsidethesidelines » Thu May 21, 2015 10:19 am
GMATGuruNY wrote: Number of options for the purple ribbon = 12. (Any of the 12 contestants.)
Number of options for the blue ribbon = 11. (Any of the 11 remaining contestants.)
Number of options for the red ribbon = 10. (Any of the 10 remaining remaining contestants.)
Number of options for the white ribbon = 9. (Any of the 9 remaining contestants.)
To combine these options, we multiply:
12*11*10*9 = [spoiler]12!/8![/spoiler].
I don't understand this at all.

When I worked through the problem, I came to (12!)/(8!4!). I understand now why that is wrong, I think, but I still don't understand how you are getting to your bottom line.

I understand using the slot method for the four ribbons, so the number of ways to award the four ribbons is 12*11*10*9, so there are 11,880 ways to award the four ribbons among the twelve individuals. Now, I understand that number must be reduced (i.e. divided) to account for the overlap, but why shouldn't it be divided by 4!, i.e. the different number of ways that the four ribbons can be awarded to the four recipients? I also don't understand how you go from 12*10*11*10 -- which I get -- to 12!/8!.

Finally, the OA, how is that correct? I understand 4! being the denominator, but why is 8! the numerator? Why would the total number of possible ways be equal to the factorial of the number of individuals who are not awarded a ribbon? To me, the correct answer should be (12*11*10*9)/(4!), but obviously I'm missing something.

Again, I just don't get this at all. Any help would be greatly appreciated.
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by [email protected] » Thu May 21, 2015 3:11 pm
Hi outsidethelines,

The type of "math" that you're discussing is an answer to a DIFFERENT question.

If the ribbons were NOT unique and the question asked "How many different ways are there to award 4 identical ribbons to 12 different people?", then the answer would be 12!/4!8!

Mitch's explanation is correct because each ribbon IS unique and each contestant can only win one ribbon (so once a contestant wins a ribbon, he/she cannot win another one).

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by GMATinsight » Fri May 22, 2015 3:46 am
Hi Another way to look at explanation of the question is

1) You have 12 participants and you need to give 4 ribbons to 4 top performers hence you need to SELECT 4 top performers out of 12 participants.
Selection of 4 best performers out of 12 participants can be done in 12C4 ways

2) Selected 4 best performers can be arranged for 4 different positions in 4x3x2x1=4! ways

i.e. total ways to give 4 ribbons to the 4 out of 12 participants = 12C4 x 4! ways

i.e. 12!/8! ways
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by outsidethesidelines » Fri May 22, 2015 6:22 am
There has to just be something I'm missing here. The original post indicates the OA is 8!/4! (i.e. 1680 ways), yet Mitch's post states the correct answer is 12!8! (i.e. 11,800 ways). What am I missing?
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by outsidethesidelines » Fri May 22, 2015 6:29 am
[email protected] wrote:Mitch's explanation is correct because each ribbon IS unique and each contestant can only win one ribbon (so once a contestant wins a ribbon, he/she cannot win another one).
Hi Rich,

I get how my answer (12!/[8!4!]) is wrong, and I understand my mistake in coming to that (and clearly you are right in your explanation) as well.

Don't get me wrong, I'm not doubting for a second that you and Mitch are right, I just don't understood how you reach that number.
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by GMATGuruNY » Fri May 22, 2015 6:41 am
outsidethesidelines wrote:There has to just be something I'm missing here. The original post indicates the OA is 8!/4! (i.e. 1680 ways), yet Mitch's post states the correct answer is 12!8! (i.e. 11,800 ways). What am I missing?
The OA posted above is wrong.
In all likelihood, the OP transcribed the OA incorrectly.
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by GMATGuruNY » Fri May 22, 2015 6:47 am
outsidethesidelines wrote:
[email protected] wrote:Mitch's explanation is correct because each ribbon IS unique and each contestant can only win one ribbon (so once a contestant wins a ribbon, he/she cannot win another one).
Hi Rich,

I get how my answer (12!/[8!4!]) is wrong, and I understand my mistake in coming to that (and clearly you are right in your explanation) as well.

Don't get me wrong, I'm not doubting for a second that you and Mitch are right, I just don't understood how you reach that number.
A permutation is an ARRANGEMENT.
The problem above asks us to count PERMUTATIONS: the number of ways that 4 of 12 contestants can be ARRANGED as prizewinners.
The number of ways to arrange k objects from a set of n objects = n!/(n-k)!.
Thus, the number of ways to arrange 4 of 12 contestants = 12!/(12-4)! = 12!/8!.
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by outsidethesidelines » Fri May 22, 2015 7:03 am
GMATGuruNY wrote:
outsidethesidelines wrote:There has to just be something I'm missing here. The original post indicates the OA is 8!/4! (i.e. 1680 ways), yet Mitch's post states the correct answer is 12!8! (i.e. 11,800 ways). What am I missing?
The OA posted above is wrong.
In all likelihood, the OP transcribed the OA incorrectly.
Haha, thanks for the confirmation, I was afraid I was losing my mind at that point.

Now, regarding your answer, I still need a further explanation. How do you go from 12*11*10*9 to 12!/8!? I'm assuming the 8! comes from the fact that 8 contestants (i.e. those not receiving a ribbon) are identical, correct? Even so, how is the jump made from 12*11*10*9 to 12!?

Many thanks in advance for all of your help.
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by outsidethesidelines » Fri May 22, 2015 7:08 am
GMATGuruNY wrote:The number of ways to arrange k objects from a set of n objects = n!/(n-k)!.
Thus, the number of ways to arrange 4 of 12 contestants = 12!/(12-4)! = 12!/8!.
Aha, now I've got you. Please ignore my reply above, and many thanks for all of the help!
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by Brent@GMATPrepNow » Fri May 22, 2015 8:57 am
GMATGuruNY wrote:
gmat_winter wrote:Twelve contestants at the county fair have entered their cakes to be judged in the cake decorating competition. A purple ribbon, blue ribbon, red ribbon, and white ribbon will be given to the first, second, third, and fourth place competitors, respectively. How many different ways are there to award the four ribbons to the contestants?
Number of options for the purple ribbon = 12. (Any of the 12 contestants.)
Number of options for the blue ribbon = 11. (Any of the 11 remaining contestants.)
Number of options for the red ribbon = 10. (Any of the 10 remaining remaining contestants.)
Number of options for the white ribbon = 9. (Any of the 9 remaining contestants.)
To combine these options, we multiply:
12*11*10*9 = [spoiler]12!/8![/spoiler].
I'd like to point out that Mitch's solution applies something called the Fundamental Counting Principle (FCP).
The FCP can be used to solve the MAJORITY of counting questions on the GMAT.

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by Amrabdelnaby » Wed Nov 18, 2015 12:09 pm
Why not 12C4?
GMATGuruNY wrote:
gmat_winter wrote:Twelve contestants at the county fair have entered their cakes to be judged in the cake decorating competition. A purple ribbon, blue ribbon, red ribbon, and white ribbon will be given to the first, second, third, and fourth place competitors, respectively. How many different ways are there to award the four ribbons to the contestants?
Number of options for the purple ribbon = 12. (Any of the 12 contestants.)
Number of options for the blue ribbon = 11. (Any of the 11 remaining contestants.)
Number of options for the red ribbon = 10. (Any of the 10 remaining remaining contestants.)
Number of options for the white ribbon = 9. (Any of the 9 remaining contestants.)
To combine these options, we multiply:
12*11*10*9 = [spoiler]12!/8![/spoiler].
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by [email protected] » Thu Nov 19, 2015 9:43 am
Hi Amrabdelnaby,

That question is addressed in the discussion above. Since the 4 ribbons are unique, and each contestant can win just one ribbon, we're dealing with a permutation.

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