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There are 5 couples(5 husband and 5 wives) in a party.Three

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There are 5 couples(5 husband and 5 wives) in a party.Three  [#permalink] New post 09 Oct 2003, 16:01
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1. There are 5 couples(5 husband and 5 wives) in a party.Three
persons are chosen in such a way that none of three are husband-wife
pair.How many such combinations are possible ?
Intern
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 [#permalink] New post 10 Oct 2003, 05:17
The no. of combinations is 80

5C2x3C1x2+5C3x2
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Re: PS : Probability (Husband/Wife) [#permalink] New post 10 Oct 2003, 06:01
Let me give it a shot.

There are 4 combinations in all.

1. MMM - 5C3
2. WWW - 5C3
3. MWM - 5C2 * 3C1 (2 men can be picked in 5C2 ways and the woman has to be one of 3 women as two men have been selected and we do not want their spouses to be on the list)
4. WMW - 5C2 * 3C1 (same here)

Legend: M - Man, W - Woman

1 and 2 are similar. So are 3 and 4.

So ,answer should be 2 * [ 5C3 + 5C2 * 3C1] = 80.

Amar.
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 [#permalink] New post 30 Oct 2003, 08:33
Can some one explain this please.

the total way of selecting 3 ppl from 10 is 10C3 = 120 WAYS.

Assuming that 1 copule is selected fro 2 posns,and another person for the third.

The guy or girl for the first posn can b selected in 10 ways.;for this person the corresponding husband/wife can be selected in only 1 way.The third independent person can be selected in 8 ways.So total is 10*1*8 = 80 ways.
So the number of ways in which no couple is selected is 120 - 80 = 40 ways.

why is this above reasoning wrong?Please explain.
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 [#permalink] New post 31 Oct 2003, 11:01
First person can be chosen in 10 ways, but since the result will be the same if one of a couple is chosen = 10/2

Similarly for the second person = 8/2
and third person = 6/2

Hence the total number of combinations = 5*4*3= 60 ways.

Praetorian, what is the correct answer?
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 [#permalink] New post 03 Nov 2003, 10:46
sud..

"The guy or girl for the first posn can b selected in 10 ways.;for this person the corresponding husband/wife can be selected in only 1 way.The third independent person can be selected in 8 ways.So total is 10*1*8 = 80 ways. "

The above is not correct. From 5 pairs you can pick up a pair only in 5 ways. Whether you take husband first or the wife first you will have only one pair (your calculation will count this as 2 separate combinations).

As the pair is chosen in five ways, the third person can be chosen from the remaing (10-2) persons in 8 ways. Therefore the combinations that will definitely include couple would be 8*5=40. Therefore the answer is 10C3-40=80.
  [#permalink] 03 Nov 2003, 10:46
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There are 5 couples(5 husband and 5 wives) in a party.Three

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