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# There are 5 married couples and a group of three is to be

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Manager
Joined: 10 Dec 2005
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There are 5 married couples and a group of three is to be [#permalink]

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20 Dec 2005, 20:48
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There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.
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JAI HIND!

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Senior Manager
Joined: 05 Oct 2005
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20 Dec 2005, 21:07
JAI HIND wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.

I'm not too good at perm/comb, but here goes:

10 ways to pick the first person, 8 ways to pick the second (b/c the 1st person's spouse must be excluded), and 6 ways to pick the 3rd (b/c the 2nd person's spouse must be excluded).

Therefore,

(10*8*6)/3! since order does not matter.

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Director
Joined: 14 Sep 2005
Posts: 984

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Location: South Korea

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20 Dec 2005, 21:14
Total number of cases to pick 3 = 10C3 = 120
Number of cases to pick 3 including one couple = 5C1 * 8C1 = 40

120 - 40 = 80

I guess yb's approach is way better.
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Auge um Auge, Zahn um Zahn !

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SVP
Joined: 14 Dec 2004
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21 Dec 2005, 12:21
cool_jonny009 wrote:
yb wrote:
JAI HIND wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.

I'm not too good at perm/comb, but here goes:

10 ways to pick the first person, 8 ways to pick the second (b/c the 1st person's spouse must be excluded), and 6 ways to pick the 3rd (b/c the 2nd person's spouse must be excluded).

Therefore,

(10*8*6)/3! since order does not matter.

Can you explain why you divided by 3!

You divide by 3! in order to get rid of same group of 3 fixed people arranged in different order & keep only one. i.e. X Y Z is actually one group but without division by 3!, all possible combinations of X Y Z will be treated as different groups.

In other words, (10*8*6) gives you total number of possible groups but this is permutation but actually we need combination, so divide by 3!

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Senior Manager
Joined: 05 Oct 2005
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21 Dec 2005, 12:30
vivek123 wrote:
cool_jonny009 wrote:
yb wrote:
JAI HIND wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.

I'm not too good at perm/comb, but here goes:

10 ways to pick the first person, 8 ways to pick the second (b/c the 1st person's spouse must be excluded), and 6 ways to pick the 3rd (b/c the 2nd person's spouse must be excluded).

Therefore,

(10*8*6)/3! since order does not matter.

Can you explain why you divided by 3!

You divide by 3! in order to get rid of same group of 3 fixed people arranged in different order & keep only one. i.e. X Y Z is actually one group but without division by 3!, all possible combinations of X Y Z will be treated as different groups.

In other words, (10*8*6) gives you total number of possible groups but this is permutation but actually we need combination, so divide by 3!

Yes, I agree. To further clarify, note that the order of XYZ does not matter.
XYZ could be :

XYZ
XZY
YXZ
YZX
ZYX
ZXY
or 3! = 3*2 = 6 ways.

Therefore, we must divide by 3! since we don't care about the order.

Kudos [?]: 6 [0], given: 0

Re: PS: P&C   [#permalink] 21 Dec 2005, 12:30
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