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Combinations - Groups

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Combinations - Groups [#permalink] New post 11 Sep 2003, 02:52
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In how many ways can 4 groups of 2 each be selected from a group of 8 Students?


Is it 8C2 * 6C2 * 4C2 * 2C2 ?

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Praetorian
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 [#permalink] New post 12 Sep 2003, 09:55
I think no. There is the doubling effect: the set of some combinations defines the set of others.

For example, we have a group of 2 people, in how many ways we can divide them? The answer is clear -- there is the only way to do so. However, if we employ combinatorics, we count twice: 2C1=2.

In your case the doubling effect is much more complicated.
I think it is [8C2*6C2*4C2*2C2]/[4!*3!*2!*1!]
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 [#permalink] New post 12 Sep 2003, 16:00
stolyar wrote:
I think no. There is the doubling effect: the set of some combinations defines the set of others.

For example, we have a group of 2 people, in how many ways we can divide them? The answer is clear -- there is the only way to do so. However, if we employ combinatorics, we count twice: 2C1=2.

In your case the doubling effect is much more complicated.
I think it is [8C2*6C2*4C2*2C2]/[4!*3!*2!*1!]


Maybe I didn't get precisely what you mean, but the number of ways you can arrange a group of 2 people is 2C2=1

Therefore I agree with Praetorian's solution. (not that I don't have any doubts)
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Re: Combinations - Groups [#permalink] New post 14 Sep 2003, 02:50
praetorian123 wrote:
In how many ways can 4 groups of 2 each be selected from a group of 8 Students?


Is it 8C2 * 6C2 * 4C2 * 2C2 ?

Thanks
Praetorian


Selecting multiple groups is EXACTLY the same as selecting one group. Let's say you are selecting 2 people from a group of 8. You are actually selected one group of 2 and one group of 6 (Which is an intuitive way of proving why 8C2 = 8C6). In the denominator of combinations with multiple groups, simply put the factorials of the number of people in each group.

2C8 = choose 2 to be IN and choosing 6 to be OUT or 8!/(2!6!)

To choose 4 groups of 2 from 8, the formula is simply 8!/(2!2!2!2!)

If you think of combinations as "adjusted" permutations, we have 8! ways to arrange 8 people, but each of the 4 pairs is an equivalent combination so we need to divide by 2^4.
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Re: Combinations - Groups [#permalink] New post 14 Sep 2003, 03:41
AkamaiBrah wrote:
Selecting multiple groups is EXACTLY the same as selecting one group. Let's say you are selecting 2 people from a group of 8. You are actually selected one group of 2 and one group of 6 (Which is an intuitive way of proving why 8C2 = 8C6). In the denominator of combinations with multiple groups, simply put the factorials of the number of people in each group.

2C8 = choose 2 to be IN and choosing 6 to be OUT or 8!/(2!6!)

To choose 4 groups of 2 from 8, the formula is simply 8!/(2!2!2!2!)

If you think of combinations as "adjusted" permutations, we have 8! ways to arrange 8 people, but each of the 4 pairs is an equivalent combination so we need to divide by 2^4.



That was awesome!
Thanks..My solution gives the same answer.
Re: Combinations - Groups   [#permalink] 14 Sep 2003, 03:41
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