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Combination ---- Help Needed

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Joined: 09 Sep 2004
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Combination ---- Help Needed [#permalink] New post 15 Sep 2004, 14:00

Just searching the archives and found this combination Problem.

In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded medals then how many groups of medals winners are possible?

Read approach of Paul and ywilfred which cleared my concepts to some extent but still have some confusion. Let me explain the confusion.

This is the combination problem as the word group has been used which means combination. Now the answer is 8C6 * 6C3 = 560. Right.
Why didn't we assume that 2 particular persons say X and Y couldn't make upto the finals, and then calculate the combinations of 3 among 6.

Now few days ago same type of problem was discussed which was :

1. A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner :

a) Where 2 of the friends are married and will not attend separately.

Here in this problem we fixed the positions of the couple first.. i.e assumed the couples are A and B and then found the remaining 3 people from 9 people.
So, we have 9C3*2C2 + 9C5 = 210 ways

My question is that why not like the semi-finalists problem we take all the combinations of couples b/c we don't know who is married with who. i.e instead of taking 2C2 for couples , why not taking all the combinations of couples formed with 12 people.

Also this forum is using XCX notation for Combinations as well as Permutations , right?

Can anybody clear this confusion?
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 [#permalink] New post 16 Sep 2004, 00:07
1. There could be some additional learning in the thread below which I thought will be a relevance and value

2. Permutations is described as nPr or n!/(n-r)! and combinations is expressed as nCr or n!/r!(n-r)!. Note in both cases r<=n.

3. The way to read the problem under discussion is that => from 8 it gets to 6 and from 6 it gets to 3. We will take the same flow for easy logic, hence it is 6 out of 8 (which is 8C6) and 3 out of 6 (which is 6C3) => totalling to 8C6*6C3 ways.

You would notice that complementary logic is best used in those situations when that is more convenient without loosing the crux of the logic.

In the couples proble, when 5 people are to selected and 2 of them are married. We dont need to really know which are these two. Why? because we can read the question simply as 2 out of 11 are couples and 9 others are not couples. If we need to choose 4 of these 11 people, I now know that I can select any 3 out of 9 and the couples OR any 5 out of the 9 and not the couples. Here the couples is the restraint added to the question and we take that into careful consideration. However, again in what order we select the 5 people also does not matter.
Joined: 09 Sep 2004
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 [#permalink] New post 16 Sep 2004, 19:46
Thanks Venksune. I got it.

But in these type of problems , one needs to think a lot. I don't know how come these problems can be solved in less than 2 minutes. Probably after practice it can be possible.
  [#permalink] 16 Sep 2004, 19:46
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