How many ways can Nicole form a committee?

Question

Nicole needs to form a committee of 3 from a group of 8 research attorneys to study possible changes to the Superior Court. If two of the attorneys are too inexperienced to serve together on the committee, how many arrangements of committees can Nicole form?

My teacher tells me to express this in terms of combinations. The question says that I need to find the total of both the number of combinations for 3 experienced attorneys and 2 experienced attorneys / 1 inexperienced attorney.

Because they asked for "how many arrangements", order does not matter, so I should use combinations. I have

3 experienced attorneys:

$\frac{6!}{3!(6-3)!}$=20 (please verify)

2 experienced / 1 inexperienced attorney:

(Please help me with the 2 experienced / 1 inexperienced attorney one. I'm so confused.)

You calculated the number of ways of selecting three experienced attorneys correctly. — N. F. Taussig, Aug 22 '15 at 17:50

score 1 · Answer 1 · answered Aug 22 '15 at 17:47

1

Hint:

For the two experienced, one inexperienced case: You need to choose two out of six experienced attorneys (how many ways?); and one out of two inexperienced attorneys (how many ways?)

answered Aug 22 '15 at 17:47

paw88789

40,402

6!/[2!(6-2)!] + 2!/[1!(2-1)!] ways? – comb Aug 22 '15 at 17:50
Multiply the results, don't add them. For each way to do the first step, there are so many ways to do the second step. – paw88789 Aug 22 '15 at 17:51
1

6!/[2!(6-2)!] * 2!/[1!(2-1)!] = 30 is this correct? – comb Aug 22 '15 at 17:52
Are we multiplying, not adding, because of the word "and" in selecting 2 experienced committees "and" 1 inexperienced committee? – comb Aug 22 '15 at 17:54
@comb The word "and" is an indication that you need to multiply. – N. F. Taussig Aug 22 '15 at 23:40

How many ways can Nicole form a committee?

1 Answers1