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# The number of possible 4-person teams that can be selected f

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Retired Moderator
Joined: 07 Jun 2014
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GRE 1: Q167 V156
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The number of possible 4-person teams that can be selected f [#permalink]  05 Aug 2018, 14:50
Expert's post
00:00

Question Stats:

75% (00:17) correct 25% (00:36) wrong based on 52 sessions
 Quantity A Quantity B The number of possible 4-person teams that can be selected from 6 people The number of possible 2-person teams that can be selected from 6 people

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Sandy
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Intern
Joined: 10 Aug 2018
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Kudos [?]: 24 [0], given: 0

Re: The number of possible 4-person teams that can be selected f [#permalink]  11 Aug 2018, 13:49
Use the combination formula:

N choose K = N!/(K!(N-K)!)

6 choose 4 = 6!/(4!(6-4)!)=6!/(4!2!)

6 choose 2 = 6!/(2!(6-2)!)=6!/(2!4!)

They are equal. It may be helpful to note the symmetry. If the number of people we're choosing from the same sized groups are equally far from the size of the group and the number 0, then the combinations are equal.

Some examples:

10 choose 3 = 10 choose 7

20 choose 5 = 20 choose 15

100 choose 20 = 100 choose 80
Retired Moderator
Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 173

Kudos [?]: 2977 [1] , given: 394

Re: The number of possible 4-person teams that can be selected f [#permalink]  23 Aug 2018, 07:37
1
KUDOS
Expert's post
Explanation

This is a classic combinatorics problem in which order doesn’t matter—that is, picking Javier and Sonya is the same as picking Sonya and Javier. A person is either on the team or not. Use the standard “order doesn’t matter” formula:

For Quantity A:

$$\frac{6!}{2! \times 4!}$$

For Quantity B:

$$\frac{6!}{4! \times 2!}$$

The two quantities are equal. Note that it is not actually necessary to reduce each quantity. The factorials are the same in each, so the resulting quantities must be equal.

This will always work—when order doesn’t matter, the number of ways to pick 4 and leave out 2 is the same as the number of ways to pick 2 and leave out 4. Either way, it’s one group of 4 and one group of 2. What actually happens to those groups (getting picked, not getting picked, getting a prize, losing a contest, etc.) is irrelevant to the ultimate solution.
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Re: The number of possible 4-person teams that can be selected f   [#permalink] 23 Aug 2018, 07:37
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