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Permutation not my fav as well... There are 5 men and 3

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karovd
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Permutation not my fav as well... There are 5 men and 3 [#permalink] New post   15 Dec 2004, 12:26
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Permutation not my fav as well... :cry:

There are 5 men and 3 women. 3 persons are picked to form a team, and there must be at least one man in the team. So how many different ways are there to pick these teams?



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Re: Permutation not my fav as well... There are 5 men and 3 [#permalink] New post   15 Dec 2004, 13:05
at least 1 man means a team with 1 man, 2 men, or three men.

Since the order doesn't seem to matter, it's a combinations question.

1 man would look like:

5x(3x2/2)=15

2 men would be:
(5x4/2)x3=30

3 men would be:
(5x4x3)/(3x2)=10

all together, we've got 55 teams.

OR YOU COULD DO THIS

Since we need at least 1 man, you could find the total number of all the teams, and the total number that have no men (ie, all women) and then subtract.

all together there are 8 people:

(8x7x6)/(3x2)=56

No men (all women):
(3x2x1)/(3x2)=1

56-1=55
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Re: Permutation not my fav as well... There are 5 men and 3 [#permalink] New post   15 Dec 2004, 19:50
This approach sounds very logical. But the OA is 46, can't really figure out why...

:roll:
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Re: Permutation not my fav as well... There are 5 men and 3 [#permalink] New post   19 Dec 2004, 13:10
What is wrong with this approach:


first select 1 man. it can be done in 5C1 = 5 ways.
now there are 7 people left (4 M and 3 W) .. select any 2 out of the 7, it can be done is 7C2 = 21 ways

so the total number is 21 * 5 = 105.
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Re: Permutation not my fav as well... There are 5 men and 3 [#permalink] New post   21 Dec 2004, 13:49
I did it the same way as Target 780. Why would it matter whether the other two people are men or women? The only constraint is to have at least one man, which can be done in 5 ways. After that you are picking a group of 2 without regard to order from 7 people. So shouldn't the answer be 5* (7c2)=105?
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Re: Permutation not my fav as well... There are 5 men and 3 [#permalink] New post   21 Dec 2004, 20:41
target780 wrote:
What is wrong with this approach:


first select 1 man. it can be done in 5C1 = 5 ways.
now there are 7 people left (4 M and 3 W) .. select any 2 out of the 7, it can be done is 7C2 = 21 ways

so the total number is 21 * 5 = 105.


This is a conundrum. I've been trying to work out why your way doesn't work, and I know why, but I don't know if I can explain it well.

5x7x6/2 assumes that the first space is only one guy, and the other spaces can be anything, thereby encompassing all other sorts of combos, with women only, some men and some women, and no women, just men. But what it also does is it allows doubling of some combinations. For example, imagine that we have our five men a, b, c, d, and e, and our 3 women f, g, and h.

Now here is one combination of 2 men and 1 woman:

b, d, g

Another combination would be

d, b, g

But those are the same thing, and should only be counted once. Using your method, they are counted twice.

In the correct method, we would figure out how many ways 2 men and 1 woman can come together, and we would automatically elminate the second possibility.

That's not the most complete answer, but I hope it sheds some light on things. I actually drew a very involved diagram to understand it all - the tree diagram. If you do the same, it should make sense.
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Re: Permutation not my fav as well... There are 5 men and 3 [#permalink] New post   22 Dec 2004, 10:14
The answer shoud be 55 only.

You can select 3 men in the following ways.

Case 1 : 1 Men 2 Women
5C1 * 3C2 = 5 * 3 = 15

Case 2 : 2 Men 1 Woman
5C2 * 3C1 = 10 * 3 = 30

Case 3 : 3 Men 0 Women
5C3 = 10
Total 55 ways

sleek
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sleek04
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Re: Permutation not my fav as well... There are 5 men and 3 [#permalink] New post   22 Dec 2004, 10:15
The answer shoud be 55 only.

You can select 3 men in the following ways.

Case 1 : 1 Men 2 Women
5C1 * 3C2 = 5 * 3 = 15

Case 2 : 2 Men 1 Woman
5C2 * 3C1 = 10 * 3 = 30

Case 3 : 3 Men 0 Women
5C3 = 10
Total 55 ways

sleek



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Re: Permutation not my fav as well... There are 5 men and 3 [#permalink]
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