there are 4 married couples 3 people r needed but there should not be his her spouse in group how

any three can be selected among 4 men or 4 woman so 2*4C3 ways. Also any two men out of 4 and any one woman out of remaining 2(as other 2 are wifes we are left with only to women) or vice versa( i.e., any two of 4 woman and one men from remaining 2 men) will make 2*4C2*2C1 ways. In total 2*4C3+2*4c2*2c1=32 ways
10 years agoHelpfull: Yes(34) No(6)

to select 3 couples from 4 couples is 4C3. And selected 3 couples to select one people from each couples is 2c1. So number of possible group is 4c3*2c1*2c1*2c1
10 years agoHelpfull: Yes(19) No(10)
first of all 1w with other three man total combination is=12
similarly,1m with other three woman total combination 12
and individual combination is 4c3+4c3=8
finally total combination is 32.
10 years agoHelpfull: Yes(11) No(1)
192..
8c1*6c1*4c*
10 years agoHelpfull: Yes(9) No(17)
4c3+4c3+4c2*2c2*3!*2
if only men are selected, if only ladies are selected, if 2 are men and 2 are ladies(this could be in 3! ways n 2 is multiplied in reverse case)
10 years agoHelpfull: Yes(3) No(3)
4*2c1*2c1*2c1
10 years agoHelpfull: Yes(3) No(1)
2 sets possible (M,M,M,M) or (w,W,W,W)=4c3+4c3=8
10 years agoHelpfull: Yes(2) No(13)
total no. of people = 8
one people can be chosen from them in 8C1 = 8 ways.
among the remaining 7 people one will be his/her spouse but remaining 6 people are not.
now from that 6 people remaining 2 member for the group can be chosen in 6C2 ways.
so, total no of ways will be 8*6C2 = 120
7 years agoHelpfull: Yes(1) No(2)
to select 3 couples from 4 couples is 4C3. And selected 3 couples to select one people from each couples is 2c1. So number of possible group is 4c3*2c1*2c1*2c1
9 years agoHelpfull: Yes(0) No(1)

there are 4 married couples 3 people r needed but there should not be his/her spouse in group .how many group are possible