1. ## Combinations and Probability

Question

A panel of 5 members is drawn from a group consisting of 2 managers, 7 assistants and 4 secretaries.

(a) How many ways can the panel be formed with at least 1 manager and 2 assistants

(b) Find the probability that the panel include 2 managers, 2 assistants and 1 secretary

(a) Should I use the following approach?

1M+2A+2S: C(2,1)*C(7,2)*C(4,2)=252
1M+3A+1S: C(2,1)*C(7,3)*C(4,1)=280
1M+4A+0S: C(2,1)*C(7,4)=70
2M+2A+1S: C(2,2)*C(7,2)*C(4,1)=84
2M+3A+0S: C(2,2)*C(7,3)=35

By addition, there are 721 combinations.

(b) P=2/13*1/12*7/11*6/10*4/9=14/6435

2. Rather than working out all the cases, simply choose the other 2 committee members from the remaining group of 10 people.

$\displaystyle a)$ $\displaystyle {2 \choose 1}{7 \choose 2}{10 \choose 2}$

$\displaystyle b)$ $\displaystyle \frac{{2 \choose 2}{7 \choose 2}{4 \choose 1}}{{13\choose 5}}$

3. It's.... at least 1 manager and 2 assistants
but I'm not sure if that means at least 1 manager and exactly 2 assistants
or at least 1 manager and at least 2 assistants.

Prosac's (a) is ok
In (b) you're putting in order and you need to eliminate that.

Annoying-mouse's (b) is ok
You didn't miss the exactly, but I don't think it's correct either.

4. Originally Posted by matheagle
It's.... at least 1 manager and 2 assistants
but I'm not sure if that means at least 1 manager and exactly 2 assistants
or at least 1 manager and at least 2 assistants.

Prosac's (a) is ok
In (b) you're putting in order and you need to eliminate that.

Annoying-mouse's (b) is ok
In (a) you missed the at least, you did exactly 1 ,2, 2.
Can't the $\displaystyle {10\choose 2}$ be of type manager, assistant or secretary though? And, therein, un-fixing the exactly-ness?