Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

a

OpenStudy (kropot72):

There is one way in which none of the bulbs are defective. There are 5 ways in which one of the bulbs is defective. There are 5C2 ways in which 2 of the bulbs are defective.

OpenStudy (kropot72):

There is only one way in which all of the bulbs are defective. If the 5 bulbs were numbered 1 ~5, then if exactly one bulb is defective it can be either number 1, number 2, number 3, number 4 or number 5. Therefore there are 5 ways in which one of the bulbs is defective. If 2 bulbs are defective this can happen in the 5C2 combinations of the 5 bulbs taken 2 at a times. 5C2 = 10. Therefore the number of different ways there can be at most 2 non-working light bulbs is 1 + 5 + 10 = 16 ways.

OpenStudy (kropot72):

*....taken 2 at a time.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!