There are 4 married couples, out of which, 3 poeple in a group is needed. But there should not be his or her spouse in the group. How many groups are possible?
2006-08-10 21:53:40 · 12 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
48 groups.
4 couples can be A,AA; B,BB; C,CC & D,DD.
A B C -1
A B CC -2
A BB C -3
A BB CC -4
AA B C -5
AA BB C - 6
AA BB CC -7
A B D - 8
A BB D - 9
A B DD - 10
A BB DD - 11
AA BB DD -12 & so on...
2006-08-10 22:04:03 · answer #1 · answered by evelyn_01 3 · 0⤊ 0⤋
2
2006-08-11 04:57:56 · answer #2 · answered by <<~~Da Cool X~~>> 2 · 0⤊ 0⤋
8!/6!-8
1 X 2 X 3 X 4 X 5 X 6 X 7 X 8 / 1 X 2 X 3 X 4 X 5 X 6 - 8
7 X 8 - 8
56 - 8
= 48
2006-08-11 10:24:24 · answer #3 · answered by BJ 2 · 0⤊ 0⤋
12
2006-08-11 05:19:11 · answer #4 · answered by Sandy 2 · 0⤊ 0⤋
12 groups
2006-08-11 05:08:38 · answer #5 · answered by supernova 1 · 0⤊ 0⤋
If there are 4 married couples, there are 8 people. But, to form a group we need to choose 3 people from a collection in which no spouses are present.
Let the 4 couples be represented by
{a1,b1}
{a2,b2}
{a3,b3}
{a4,b4}
where the a's are males and the b's are females.
It's pretty easy to see that if a1 is in the collection from which the group will be formed, then b1 cannot be. In fact, starting with a1, there are 2^3 ways that a collection containing no spouses may be formed.
a1 + {a2,a3,a4}
a1 + {a2,a3,b4}
a1 + {a2,b3,a4}
etc. (the triple literally 'counts' itself in binary)
Since we may start off with any of the 8 individuals, there are 8*2^3 = 64 ways to form a collection of 4 individuals in which no spouses are present.
The number of ways in which 3 people can be chose from a group of 4 is
4!/(3!(4-3)!) = 4
so the total number of ways to form a group of 3 from 4 couples without a spouse in the group is
4*64 = 256 ways
Doug
2006-08-11 05:31:07 · answer #6 · answered by doug_donaghue 7 · 0⤊ 0⤋
Let the couples be AA`,BB`,CC`,DD`
Total we have 8 persons
No. of people in one group=3
1>If all Men (A,B,C,D) then no of groups= 4C3=4
2>If all are Ladies (A`,B`,C`,D`) then no. of groups =4C3=4
3>When we combine
i)2Ladies and 1Man
(Select any of the two ladies from the four present)*(Select any one of the remaining 2 Men whose wife is not there)
=4C2*2C1=12
ii)2Men and 1Lady
(Select any of the two Men from the four present)*(Select any one of the remaining 2 ladies whose husband is not there)
=4C2*2C1=12
Hence total groups of 3 are 4+4+12+12=32
2006-08-11 07:03:24 · answer #7 · answered by awesomeash 2 · 0⤊ 0⤋
assume there are 4 pairs of couples
the total no of ways of selecting groups of three is
4C3*2*2*2=32
2006-08-11 05:15:07 · answer #8 · answered by keerthan 2 · 0⤊ 0⤋
6C2 x 2 = 30
2006-08-11 08:33:27 · answer #9 · answered by Masquerade 2 · 0⤊ 0⤋
either 3 girls or 3 guys... so 2 groups........
2006-08-11 05:03:44 · answer #10 · answered by doable_rods 5 · 0⤊ 0⤋