SOLUTION: A 6-member committee is to be formed in a hospital unit that has 8 RNs and 4 LPNs. If the committee must have at least 3 LPNs, how many different committees could be formed given t

Question 1192277: A 6-member committee is to be formed in a hospital unit that has 8 RNs and 4 LPNs. If the committee must have at least 3 LPNs, how many different committees could be formed given these conditions?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Saying "at least 3 LPNs" means "3 LPNs or more".

We'll break things down into two mutually exclusive cases
Case A: Having exactly 3 LPNs
Case B: Having exactly 4 LPNs

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Case A: Having exactly 3 LPNs

There are 4 ways to pick the 3 LPNs, because it's equivalent to 4 ways to not pick a particular unlucky LPN.

For the RNs, we have n = 8 of them total and we select r = 3 of them where order doesn't matter.
Order doesn't matter because each person on the committee does not outrank any other, and no seats are labeled or special (eg: we don't have a chairman or president)

Since order doesn't matter, we use the nCr combination formula.
n C r = (n!)/(r!(n-r)!)
8 C 3 = (8!)/(3!*(8-3)!)
8 C 3 = (8!)/(3!*5!)
8 C 3 = (8*7*6*5!)/(3!*5!)
8 C 3 = (8*7*6)/(3!)
8 C 3 = (8*7*6)/(3*2*1)
8 C 3 = (336)/(6)
8 C 3 = 56
There are 56 ways to pick 3 RNs from a pool of 8.

We found there are:
4 ways to pick the three LPNs
56 ways to pick the three RNs

Therefore, we have 4*56 = 224 ways to pick all six people to satisfy the criteria for case A.

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Case B: Having exactly 4 LPNs

There's only one way to select all four LPNs since we have four total to pick from.
The order of selection doesn't matter (refer to case A above).

We have r = 6-4 = 2 slots left for the n = 8 RNs to pick from.

n C r = (n!)/(r!(n-r)!)
8 C 2 = (8!)/(2!*(8-2)!)
8 C 2 = (8!)/(2!*6!)
8 C 2 = (8*7*6!)/(2!*6!)
8 C 2 = (8*7)/(2!)
8 C 2 = (8*7)/(2*1)
8 C 2 = (56)/(2)
8 C 2 = 28

There are 28 ways to pick the two RNs.

So there are 1*28 = 28 ways to pick all six people for case B.

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The summary values of cases A and B are

Case A) 224 ways to pick 3 LPNs and 3 RNs
Case B) 28 ways to pick 4 LPNs and 2 RNs

Cases A and B are mutually exclusive.
There is no overlap between them.
This allows us to add the counts found.
A+B = 224+28 = 252 is the final answer

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