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I'll start off by admitting that this is a HW problem. However, my main goal is to find out what kind of problem this is classified as and what concepts I need to learn to be able to solve similar ones. I'm not just looking for an answer (although I would be glad to get one); I need help by being pointed in the right direction. Links to relevant articles and readings would be appreciated.


A shipment of 12 X-Ray machines includes 4 that are defective. In how many ways can a hospital purchase 5 of these machines and receive at least 2 of the defective machines?

I can only get as far as figuring out that there are 792 different ways to purchase 5 machines from the shipment of 12 (12 choose 5 is 792).

I realize this might be very easy for some people, but all I need is a little help (links and explanations is what I'm after, not just an answer). Thanks in advance!

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I found two similar questions online but their answers are not too helpful 1st link 2nd link –  Cashew Sep 15 '12 at 19:01
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1 Answer 1

up vote 1 down vote accepted

Suppose I want to pick $x$ bad machines, and $y$ good machines for a total of $x+y$ machines. First I pick the $x$ bad ones by choosing from the set of bad ones, so there's $\binom{4}{x}$ ways of picking the bad ones and $\binom{8}{y}$ ways to pick the good ones. Multiply the two to get the total number of ways. Since you want at least two defective ones, sum all the possible cases from 2 to 4.

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Ok, so it would be the sum of (4Cx)*(12Cy) as I vary x from 2 to 4, right? But, do I also vary y so as to keep the sum x+y=5 ? like this: (4C2)*(12C3) + (4C3)*(12C2) + (4C4)*(12C1) is this correct? –  Cashew Sep 15 '12 at 19:34
Do you happen to know what the name of the "rule" or concept you used to answer the question? I would very much like to read more about it... Thanks again! –  Cashew Sep 15 '12 at 19:42
wait, wouldn't I use (8Cy) instead of (12Cy) because there are only 8 "good" machines to pick from? –  Cashew Sep 17 '12 at 20:51
Ah sorry, I thought there were 12+4 machines. Thanks for the correction! –  Alex R. Sep 18 '12 at 1:17
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