Question

A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different.)

Answer 1

Answer:

There are 100 different groups that can be formed.

Step-by-step explanation:

We will define:

• $N_1$=The number of different groups of 3 partners that can be formed in which exactly one member of the group is a senior partner.
• $N_2$=The number of different groups of 3 partners that can be formed in which exactly two members of the group are senior partners.
• $N_3$=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 $N_1+N_2+N_3$, then we will compute them.

• $N_1$. 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 ${4\choose 1}{6\choose 2}=\frac{4!}{3!}\frac{6!}{4!2!} =\frac{6!}{3!2!} =60.$

• $N_2$. 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 ${4\choose 2}{6\choose 1}=\frac{4!}{2!2!}\frac{6!}{5!} =36.$

• $N_3$. 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 ${4\choose 3}{6\choose 0}=\frac{4!}{3!}=4.$

Therefore:

$N_1+N_2+N_3=60+36+4=100.$