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

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

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16 Oct 2004, 12:46
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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?

How many different signals can be transmitted by hoisting 3 red, 4 yellow and 2 blue flags on a pole, assuming that in transmitting a signal all nine flags are to be used?
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16 Oct 2004, 13:07
1. All possible combn - (no of married couple) X (no of choice for third person)

10C3 - (5C1 X 8C1) = 120 - 40 = 80

2. 9 ! / (3! x 4 ! x 2 !) = 1260
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17 Oct 2004, 07:40
1. let the 5 couples be represented as
A(H,W); B(H,W); C(H,W); D(H,W); E(H,W); where H represent Husband and W wife.

now out of these 5 grps, u have to select 3; only then u will get a group of mutually unmarried ppl; so that can be done in 5C3 ways. However from within a grp (A,b,c,d e) you can select either H or W. and that can be done in 2C1 ways.
SO the total no of ways = 5C3 * 2C1 = 10 * 2 = 20.

2. its a question of no of permutations of 9 things where 3 are of one type, 4 of the other and 2 of the third type and that is 9! / (3! * 4! * 2!) .... a huge number, not worth the type spending on calculation...
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17 Oct 2004, 08:00
1) I get 120

Total number of 3 you can choice from a group of 10 (5 couples) = 10 C 3
Total number of groups of 3 where a H and W are in the same group. 8C1 * 5 (You can think of the HW as one, so out of the eight remaining choose 1, since there are 5 couples, the HW and be any one of the five)

10C3 - (8C1 * 5) = 120 - 40 - 80

2) Total 9 flags, = 9! Divide by non-unique (4!*3!*2!) = 9!/(4!*3!*2!) = 1260
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17 Oct 2004, 12:14
For 1 ... I'm going with

M-M-M groups
5*4*3/3! = 10

+

F-F-F groups
5*4*3/3! = 10

+

M-M-F groups
5*4*3/2! = 30

+

F-F-M groups
5*4*3/2! = 30

Total is 80 ...

Additionally, I think that the previous post from Target780 is correct except that it should be
5C3 * 2^3 = 80
17 Oct 2004, 12:14
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