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The number of possible 4-person teams that can be selected f
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05 Aug 2018, 14:50
05 Aug 2018, 14:50
Expert Reply
00:00
Question Stats:
76% (00:26) correct
23% (00:43) wrong based on 84 sessions
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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.
Re: The number of possible 4-person teams that can be selected f
[#permalink]
11 Aug 2018, 13:49
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
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
Re: The number of possible 4-person teams that can be selected f
[#permalink]
23 Aug 2018, 07:37
23 Aug 2018, 07:37
1
Expert Reply
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.
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.
