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JP, you said you had a problem like this one, so I thought

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JP, you said you had a problem like this one, so I thought [#permalink] New post 23 Jul 2003, 08:10
JP, you said you had a problem like this one, so I thought I'd throw two out there...

A group consists of 10 men and 10 women. How many pairs can be made to contain at least one woman?

A group consists of 17 men and 22 women. How many pairs can be made to contain at least one woman?
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 [#permalink] New post 23 Jul 2003, 08:18
(1) A group consists of 10 men and 10 women = 20 people.

20C2 is the total number of pairs = 190
10C2 is the number of pairs consisting of two men = 45

190-45=145

(2) the same approach
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 [#permalink] New post 24 Jul 2003, 00:51
so for part 2:

39C2=741- total pairs
17C2=136- men pairs

Then, answer is 605.

Wow that's a lot of different combination of girls! It would make an interesting blind-date TV show.
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another approach [#permalink] New post 25 Jul 2003, 01:30
Stolyar's approach is the most concise and clear. However, if on the official test this question appears last in the maths section and you happened to have 15 minutes left :shock: , check yourself as follows:

At least one means one or more (i.e. in our case favourable conditions are: one woman in a pair, two women in a pair). Thus,

P (one woman in a pair (and one man)) = C(10,1)*C(10,1)=100.
We use multiplication because both outcomes are parts of one set.

P (two women) = C(10,2)=45.

Add the outcomes - 100+45=145

Nyammy...
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Respect,

KL

another approach   [#permalink] 25 Jul 2003, 01:30
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