SOLUTION: The shipment of 10 television sets contain 3 defectives sets . In hw many ways can a dealer purchase 4 of these sets and recieve at least 2 of the defectives sets

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Question 887210: The shipment of 10 television sets contain 3 defectives sets . In hw many ways can a dealer purchase 4 of these sets and recieve at least 2 of the defectives sets
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the 10 tv sets include 7 good and 3 defective
to get at least 2 defective in a set of 4, the dealer needs to get sets of 4 tv's that include exactly 2 defectives each and sets of 4 tv's that include exactly 3 defectives each.

each set of 2 defectives will include 2 good and 2 defectives.
each set of 3 defectives will include 1 good and 3 defectives.

the number of ways you can get a set of 2 defectives and 2 good is equal to:
3C2 * 7C2

the number of ways you can get a set of 3 defectives and 1 good is equal to:
3C3 * 7C1

3C2 * 7C2 = 3 * 21 = 63

3C3 * 7C1 = 1 * 7 = 7

total number of ways to get sets of 4 that include at least 2 defectives each is equal to 63 + 7 = 70.

nCx formula is equal to n! / (x! * (n-x)!

for example:

7C2 = 7! / (2! * 5!) = (7*6*5*4*3*2*1) / (2*1*5*4*3*2*1) = 21