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Can someone explain me the answer to this qestion:
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? (Ans. 80)
I thought it is: (10!/(3!*7!)-(5!/(3!*2!))= 110.
Thanks!
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? (Ans. 80)
I thought it is: (10!/(3!*7!)-(5!/(3!*2!))= 110.
Thanks!
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17 Jul 2007, 09:03
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My approach:
Total combinations of choosing 3 out of 10 = 10C3 = 120
Now calculate the number of combinations, which have husband and wife together,
Such combinations will be :
H1W1X - 8 ways
Hence for other couples
So total such ways = 5*8 = 40
Hence the number of combinations, which do not have husband and wife together = 120-40 =80
Total combinations of choosing 3 out of 10 = 10C3 = 120
Now calculate the number of combinations, which have husband and wife together,
Such combinations will be :
H1W1X - 8 ways
Hence for other couples
So total such ways = 5*8 = 40
Hence the number of combinations, which do not have husband and wife together = 120-40 =80
18 Jul 2007, 00:46
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Sorry, but I do not understand this step:
Now calculate the number of combinations, which have husband and wife together,
Such combinations will be :
H1W1X - 8 ways
For me it is:
H/W/X
H/X/W
W/H/X
W/X/H
X/H/W
X/W/H
so, I can only see 6 ways.
(Probably it is so easy, but I can´t see it.)
Thanks a lot!
Now calculate the number of combinations, which have husband and wife together,
Such combinations will be :
H1W1X - 8 ways
For me it is:
H/W/X
H/X/W
W/H/X
W/X/H
X/H/W
X/W/H
so, I can only see 6 ways.
(Probably it is so easy, but I can´t see it.)
Thanks a lot!
