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Difficult Math Question #67 - Combinations

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Question #67 - Combinations

Post Wed Nov 29, 2006 2:49 pm
A shipment of 10 TV sets includes 3 that are defective. In how many ways can a hotel purchase 4 of these sets and receive at least two of the defective sets?

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800guy Master | Next Rank: 500 Posts Default Avatar
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Post Fri Dec 01, 2006 5:26 pm
OA:

There are 10 TV sets; we have to choose 4 at a time. So we can do that by 10C4 ways. We have 7 good TV's and 3 defective.

Now we have to choose 4 TV sets with at least 2 defective. We can do that by

2 defective 2 good
3 defective 1 good

That stands to 3C2*7C2 + 3C3*7C1 (shows the count)

If they had asked probability for the same question then

3C2*7C2 + 3C3*7C1 / 10C4.

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kulksnikhil Senior | Next Rank: 100 Posts Default Avatar
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Post Fri Dec 01, 2006 3:55 am
800guy wrote:
A shipment of 10 TV sets includes 3 that are defective. In how many ways can a hotel purchase 4 of these sets and receive at least two of the defective sets?
Possible ways

1> All 3 defective and 1 correct
2> 2 defective and 2 correct


1> 1 * 7 = 7
2> 3C2 * 7C2 = 6 * 21 = 126

total= 126 + 7 = 133

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800guy Master | Next Rank: 500 Posts Default Avatar
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Post Fri Dec 01, 2006 5:26 pm
OA:

There are 10 TV sets; we have to choose 4 at a time. So we can do that by 10C4 ways. We have 7 good TV's and 3 defective.

Now we have to choose 4 TV sets with at least 2 defective. We can do that by

2 defective 2 good
3 defective 1 good

That stands to 3C2*7C2 + 3C3*7C1 (shows the count)

If they had asked probability for the same question then

3C2*7C2 + 3C3*7C1 / 10C4.

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