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Re: Combinations - Groups [#permalink]
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]
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.



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