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Combos

by Jinglander » Sun Aug 08, 2010 4:40 pm
9 players are trying out for the team. 5 will be selected. If 6 are gaurds and 3 forwards how many different teams would have 3 guards and 2 forward
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by HPengineer » Sun Aug 08, 2010 5:25 pm
Could you use slot method??

6 x 5 x 4 for the guards 3 x 2 for the forwards then divide by 3! and 2!....

Probably way wrong but figured i would give it a shot...
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by Stuart@KaplanGMAT » Sun Aug 08, 2010 5:38 pm
HPengineer wrote:Could you use slot method??

6 x 5 x 4 for the guards 3 x 2 for the forwards then divide by 3! and 2!....

Probably way wrong but figured i would give it a shot...
You could indeed!

So: 6*5*4*3*2/3!2! = 6*5*4*3*2/3*2*2 = 20*3 = 60

Of course, that's just the combinations formula:

6C3 * 3C2 = (6!/3!3!) * (3!/2!1) = (6*5*4/3*2) * (3) = 20 * 3 = 60

For the original poster - instead of repeating the question in your own words, please copy it exactly. Also, please provide the source and the answer choices.

Thanks!
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by likithae » Mon Aug 09, 2010 4:31 am
Jinglander wrote:9 players are trying out for the team. 5 will be selected. If 6 are gaurds and 3 forwards how many different teams would have 3 guards and 2 forward
total number of players =9

in that 6 are guards ,3 are forwards

for team needed=5 members

3 guards+2 forward=5 members=1 team

the number of teams can be formed =6C3*3C2=60
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