SOLUTION: a computer store receives a shipment of 24 monitors, including 5 defective and 19 good monitors. Three monitors are chosen at random. How many selections of 3 are possible? How

Question 1083597: a computer store receives a shipment of 24 monitors, including 5 defective and 19 good monitors. Three monitors are chosen at random.
How many selections of 3 are possible?
How many selections will contain NO defective monitors out of 3?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Problem:
A computer store receives a shipment of 24 monitors, including 5 defective and 19 good monitors. Three monitors are chosen at random.

Questions:
Part A) How many selections of 3 are possible?
Part B) How many selections will contain NO defective monitors out of 3?
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Answers:

Answer for part A): 2024
Answer for part B): 969

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Explanation:

Part A)

We have 24 monintors total. The pool size is n = 24.
When I use the term "pool", I mean the set in which we can make a selection from.

We have 3 slots to fill. This means r = 3.
Order does NOT matter. Why not?
Because all that matters is the grouping overall.
A group like {A,B,C} is the same as {C,B,A}
The only thing important really is the condition of the monitor (whether it's defective or non-defective).

Use the combination formula to get

n C r = (n!)/(r!(n-r)!)
24 C 3 = (24!)/(3!*(24-3)!)
24 C 3 = (24!)/(3!*21!)
24 C 3 = (24*23*22*21!)/(3!*21!)
24 C 3 = (24*23*22)/(3!)
24 C 3 = (24*23*22)/(3*2*1)
24 C 3 = (12144)/(6)
24 C 3 = 2024

which is the answer to part A.

An alternative is to think of having 24 choices for slot A
23 choices for slot B, and 22 choices for slot C

That makes 24*23*22 = 12144 different permutations.

Because order does NOT matter, we divide by 3! = 3*2*1 = 6 because there are 6 ways to order any single group of three.

Doing so yields: 12144/6 = 2024
leading to the same answer.
This alternative method of thinking is shown in steps 5 & 6 of the combination formula method above.

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Part B)

The pool of possible choices goes from 24 to 19 since we're only considering the non-defective monitors now.
So n = 19 for part B.

We still have 3 slots to fill. So r = 3.
Like before, order isn't important.

Use the combination formula.

n C r = (n!)/(r!(n-r)!)
19 C 3 = (19!)/(3!*(19-3)!)
19 C 3 = (19!)/(3!*16!)
19 C 3 = (19*18*17*16!)/(3!*16!)
19 C 3 = (19*18*17)/(3!)
19 C 3 = (19*18*17)/(3*2*1)
19 C 3 = (5814)/(6)
19 C 3 = 969

which is the answer to part B

Side Note:
Subtract the results from part A) and part B) to get
2024-969 = 1055
This value of 1055 is the number of ways to select a trio of monitors where at least one is defective.
"At least one" means "one or more".
This bit of info is optional but may come in handy down the line.

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