Answer:
1) 35 distinct groups can be formed.
2) 18 distinct groups can be formed containing 2 men and 1 woman.
Step-by-step explanation:
The no of groups of 3 members that can be chosen from 7 members equals no of combinations of 3 members that can be formed from 7 members.
Thus no of groups =
[tex]n=\binom{7}{3}=\frac{7!}{(7-3)!\times 3!}=35[/tex]
thus 35 distinct groups can be formed.
Part b)
Now since the condition is that we have to choose 2 men and 1 women to form the group
let A and B be men member's of group thus we have to choose 2 member's from a pool of 4 men which equals
[tex]\binom{4}{2}=\frac{4!}{(4-2)!\times 2!}=6[/tex]
Let the Woman member be C thus we have to choose one woman from a pool of 3 women hence number of ways in which it can be done equals 3.
thus the group can be formed in [tex]6\times 3=18[/tex] different ways.
