Answer:
There are 100 different groups that can be formed.
Step-by-step explanation:
We will define:
- [tex]N_1[/tex]=The number of different groups of 3 partners that can be formed in which exactly one member of the group is a senior partner.
- [tex]N_2[/tex]=The number of different groups of 3 partners that can be formed in which exactly two members of the group are senior partners.
- [tex]N_3[/tex]=The number of different groups of 3 partners that can be formed in which exactly three members of the group are senior partners.
Observe that the quantity we are looking for is [tex]N_1+N_2+N_3[/tex], then we will compute them.
- [tex]N_1[/tex]. In this case we want to choose 1 senior partner between 4 and 2 junior partners between 6. Since the choices are independent, the numbers of ways to do this is given by [tex]{4\choose 1}{6\choose 2}=\frac{4!}{3!}\frac{6!}{4!2!} =\frac{6!}{3!2!} =60.[/tex]
- [tex]N_2[/tex]. In this case we want to choose 2 senior partners between 4 and 1 junior partner between 6. Since the choices are independent, the numbers of ways to do this is given by [tex]{4\choose 2}{6\choose 1}=\frac{4!}{2!2!}\frac{6!}{5!} =36.[/tex]
- [tex]N_3[/tex]. In this case we want to choose 3 senior partners between 4 and 0 junior partners between 6. Since the choices are independent, the numbers of ways to do this is given by [tex]{4\choose 3}{6\choose 0}=\frac{4!}{3!}=4.[/tex]
Therefore:
[tex]N_1+N_2+N_3=60+36+4=100.[/tex]