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by [email protected] » Thu Jul 01, 2010 6:01 am
In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded medals, then how many groups of medal winners are possible?
A. 20
B. 56
C. 120
D. 560
E. 720
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by kvcpk » Thu Jul 01, 2010 6:33 am
[email protected] wrote:In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded medals, then how many groups of medal winners are possible?
A. 20
B. 56
C. 120
D. 560
E. 720
first 6 members need to be picked.
So 8c6 = 28

from these 6, three members need to be picked
so 6c3 = 20

both needs to happen
so 28*26 = 560
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by GMATGuruNY » Thu Jul 01, 2010 7:33 am
Be careful here. The final round is immaterial. There are 8 contestants. Couldn't any one of these 8 be among the final 3 medal winners? All we need to determine is the the number of combinations of 3 that can be made from the 8 contestants:

(8*7*6)/(1*2*3)=56.
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by kvcpk » Thu Jul 01, 2010 7:37 am
GMATGuruNY wrote:Be careful here. The final round is immaterial. There are 8 contestants. Couldn't any one of these 8 be among the final 3 medal winners? All we need to determine is the the number of combinations of 3 that can be made from the 8 contestants:

(8*7*6)/(1*2*3)=56.
Thanks for the correction Mitch!!
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by selango » Thu Jul 01, 2010 7:40 am
3 peoples to be selected from 8.

8C3=56
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by GMATGuruNY » Thu Jul 01, 2010 7:47 am
kvcpk wrote:
GMATGuruNY wrote:Be careful here. The final round is immaterial. There are 8 contestants. Couldn't any one of these 8 be among the final 3 medal winners? All we need to determine is the the number of combinations of 3 that can be made from the 8 contestants:

(8*7*6)/(1*2*3)=56.
Thanks for the correction Mitch!!
Happy to help!

Here's what the arithmetic you performed above represents:

8c6 = the number of combinations of 6 that can be made from 8 choices = 28

6c3 = the number of combinations of 3 that can be made from 6 choices = 20

28 * 20 = 560 determines the number of ways we can COMBINE the combinations of 6 with the combinations of 3. It's as if we had 28 shirts and 20 ties and we we were determining how many ways we could combine them to make an outfit.

One more piece of advice. Given that we have only 8 contestants total, 560 is a huge number of combinations. Before you choose an answer choice, always ask yourself whether it makes sense given the constraints of the problem.
Last edited by GMATGuruNY on Thu Jul 01, 2010 8:16 am, edited 1 time in total.
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by gmat1011 » Thu Jul 01, 2010 7:56 am
Just to clarify -

What would 8c6 X 6c3 = 560 signify? I mean what would the number 560 represent in mathematical terms?

Why can it not represent sequential group formations: (i) first of a group of 6 out of 8; and then (ii) a second group of 3 medal winners out of the selected 6 in (i) --- where order does not matter...

Many thanks.
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by gmat1011 » Thu Jul 01, 2010 7:57 am
Thanks Mitch --- seems you have already answered it above!
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by raunakrajan » Thu Jul 01, 2010 8:20 am
GMATGuruNY wrote:
kvcpk wrote:
GMATGuruNY wrote:Be careful here. The final round is immaterial. There are 8 contestants. Couldn't any one of these 8 be among the final 3 medal winners? All we need to determine is the the number of combinations of 3 that can be made from the 8 contestants:

(8*7*6)/(1*2*3)=56.
Thanks for the correction Mitch!!
Happy to help!

Here's what the arithmetic you performed above represents:

8c6 = the number of combinations of 6 that can be made from 8 choices = 28

6c3 = the number of combinations of 3 that can be made from 6 choices = 20

28 * 20 = 560 determines the number of ways we can COMBINE the combinations of 6 with the combinations of 3. It's as if we had 28 shirts and 20 ties and we we were determining how many ways we could combine them to make an outfit.

One more piece of advice. Given that we have only 8 contestants total, 560 is a huge number of combinations. Before you choose an answer choice, always ask yourself whether it makes sense given the constraints of the problem.
Hi Mitch,
I din quite get the Shirt, Tie analogy.
I still din't understand whats wrong with 560 even i got that answer.
pls explain
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by GMATGuruNY » Thu Jul 01, 2010 8:39 am
raunakrajan wrote:
GMATGuruNY wrote:
kvcpk wrote:
GMATGuruNY wrote:Be careful here. The final round is immaterial. There are 8 contestants. Couldn't any one of these 8 be among the final 3 medal winners? All we need to determine is the the number of combinations of 3 that can be made from the 8 contestants:

(8*7*6)/(1*2*3)=56.
Thanks for the correction Mitch!!
Happy to help!

Here's what the arithmetic you performed above represents:

8c6 = the number of combinations of 6 that can be made from 8 choices = 28

6c3 = the number of combinations of 3 that can be made from 6 choices = 20

28 * 20 = 560 determines the number of ways we can COMBINE the combinations of 6 with the combinations of 3. It's as if we had 28 shirts and 20 ties and we we were determining how many ways we could combine them to make an outfit.

One more piece of advice. Given that we have only 8 contestants total, 560 is a huge number of combinations. Before you choose an answer choice, always ask yourself whether it makes sense given the constraints of the problem.
Hi Mitch,
I din quite get the Shirt, Tie analogy.
I still din't understand whats wrong with 560 even i got that answer.
pls explain
Say you had two closets A and B. Closet A contains 8 items of clothing. Closet B has 6. Each item of clothing is of a different color, and you plan to make an outfit by picking 6 items from closet A and 3 from closet B. How many color combinations could your outfit contain?

8c6 = the number of possible 6-color combinations from closet A = 28

6c3 = the number of possible 3-color combinations from closet B = 20

28 * 20 = 560 = the number of ways to combine the 6-color combinations from closet A with the 3-color combinations from closet B. (When you're choosing from different sources -- in this case, from closet A and from closet B -- the total number of combinations = (number of choices from source 1) * (number of choices from source 2).

So your outfit could contain 560 different color combinations. (By the way, if you really do this, Lady Gaga has nothing on you.)

Does the explanation above help?
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