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In the problem "There are 12 fuses, 5 of which are blown-out. In how many ways can 4 fuses be selected such that at least 3 are blown-out?" The answer is

(5C3)(7C1) + (5C4)(7C0)

My question is why it cannot be

(5C3)(9)

when the nature of the remaining fuse does not matter.

Sabibble
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  • Let's try an easier problem: "All you have is $2$ fuses, and they are both blown-out. In how many many ways can $2$ fuses be selected so that at least $1$ is blown-out?" By your logic the answer should be $(_2C_1)(1)=2$ because there are two ways to choose one blown-out fuse, and then there's one fuse left. Well, is $2$ the right answer? Are there $2$ ways to do that? – bof Mar 13 '17 at 11:30

1 Answers1

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In at least case we include given cases and above cases also.

So here are two cases when we picked 4 -

1.) Out of 5 we have picked 3 blown and out of 7 we have picked 1 is working.

2.) Out of 5 we have picked 4 blown.

Amar
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