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# [Combinations] Double Games

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Manager
Joined: 12 May 2009
Posts: 53

Kudos [?]: 117 [0], given: 18

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17 Jul 2009, 16:54
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Hi folks,

I got stuck with the following combinations question:

"10 tennis players are available. Their coach wants to schedule double games (2 players play another 2 players). How many different games can be scheduled?"

2C10 * 2C8 = 1260. Why is my answer wrong?

Kudos [?]: 117 [0], given: 18

Manager
Joined: 03 Jul 2009
Posts: 106

Kudos [?]: 93 [1], given: 13

Location: Brazil

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18 Jul 2009, 13:35
1
KUDOS
You were almost there!

The answer is $$(2C10 * 2C8) / 2$$

You forgot to consider that when the pair A plays with pair B, is the same game when pair B plays with pair A.

Because you did not consider it, you are counting
pair A x pair B - 1 game
pair B x pair A - another different game, but in fact they are equal.

Kudos [?]: 93 [1], given: 13

Manager
Joined: 12 May 2009
Posts: 53

Kudos [?]: 117 [0], given: 18

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18 Jul 2009, 13:55
Thanks coelholds!

I sort of get what you're trying to say! But maybe you can clear up some confusion for me:

Let's say we have 10 tennis players: (A, B, C, D, E, F, G, H, I, J, L)

1) When I select 2 people out of this set, I might end up with (A, E) -- but not (E, A) since we're not talking about permutations...
2) With 2C8, I select 2 people out of 8. Now where I get confused is this: what does 8 mean in this context? Does it preclude the 2 people that I already chose under 1)? Because if it does, I do not need to divide by 2 right?

Thanks again. This is not my strong game...

Kudos [?]: 117 [0], given: 18

Manager
Joined: 03 Jul 2009
Posts: 106

Kudos [?]: 93 [1], given: 13

Location: Brazil

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18 Jul 2009, 14:57
1
KUDOS
1) Yes, you are right

2) Yes you are right in the first part. The 8 is because the 2 people precluded in 1)

You still must divide by two by two, because when you calculate using C, it already gives you (A,E) but not (E,A). However, you are multiplying and in doing this you are not using the C formula any more, and thus you are counting pairA x pairB twice.

Imagine that you have only 3 PAIRS (not players), X, Y, Z. How many games it is possible?

If you do 3*2 = 6 you are wrong, because you are counting some games twice:
X x Y
X x Z
Y x X
Y x Z
Z x X
Z x Y

As you can see, the bold ones are repeated. That is why you must divide by two. You are not using the formula any more. The formula of Combination is just use in the first part of the problem.

Now is it clear?

If not, you can just ask, ok? Sometimes it is hard to explain only through text....

Kudos [?]: 93 [1], given: 13

Manager
Joined: 12 May 2009
Posts: 53

Kudos [?]: 117 [0], given: 18

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18 Jul 2009, 15:20
coelholds! That's a very good explanation! Thanks a lot!

Life is good

Kudos [?]: 117 [0], given: 18

Manager
Joined: 27 Jun 2008
Posts: 155

Kudos [?]: 34 [0], given: 11

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22 Jul 2009, 00:57
Yeah AB vs CD is equal to CD vs AB. Divide by 2 again.

Kudos [?]: 34 [0], given: 11

Manager
Joined: 14 Nov 2008
Posts: 195

Kudos [?]: 129 [0], given: 3

Schools: Stanford...Wait, I will come!!!

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22 Jul 2009, 06:16
Thanks a lot.
Can you suggest any reference material to improve the permutation and combination?
coelholds wrote:
1) Yes, you are right

2) Yes you are right in the first part. The 8 is because the 2 people precluded in 1)

You still must divide by two by two, because when you calculate using C, it already gives you (A,E) but not (E,A). However, you are multiplying and in doing this you are not using the C formula any more, and thus you are counting pairA x pairB twice.

Imagine that you have only 3 PAIRS (not players), X, Y, Z. How many games it is possible?

If you do 3*2 = 6 you are wrong, because you are counting some games twice:
X x Y
X x Z
Y x X
Y x Z
Z x X
Z x Y

As you can see, the bold ones are repeated. That is why you must divide by two. You are not using the formula any more. The formula of Combination is just use in the first part of the problem.

Now is it clear?

If not, you can just ask, ok? Sometimes it is hard to explain only through text....

Kudos [?]: 129 [0], given: 3

Manager
Joined: 03 Jul 2009
Posts: 106

Kudos [?]: 93 [0], given: 13

Location: Brazil

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22 Jul 2009, 06:36
Sorry lgon, all the math material that I have is in Portuguese.

Kudos [?]: 93 [0], given: 13

Manager
Joined: 12 May 2009
Posts: 53

Kudos [?]: 117 [0], given: 18

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22 Jul 2009, 06:52
www.4GMAT.com offers a combinations and probablility e-book at \$6.99. Does anyone have some experience with 4GMAT ebooks? I feel that they're providing very hard questions but the written English is pretty bad sometimes (which probably confuses ppl when they try to understand the question. Plus, GMAT questions are just worded differently)

Kudos [?]: 117 [0], given: 18

Re: [Combinations] Double Games   [#permalink] 22 Jul 2009, 06:52
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