Difficult Math Question #67 - Combinations
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Possible ways800guy 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?
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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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.
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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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.
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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If we let N = normal TV and D = defective TV, there are two ways we can receive at least 2 defective TVs
Case 1: DDNN
Case 2: DDDN
(DDDD can't happen because there are only 3 defective TVs)
Therefore, the total number of ways that the hotel can receive at least 2 defective TVs is
(3C2)*(7C2) + (3C3)(7C1) = 70 ways
Case 1: DDNN
Case 2: DDDN
(DDDD can't happen because there are only 3 defective TVs)
Therefore, the total number of ways that the hotel can receive at least 2 defective TVs is
(3C2)*(7C2) + (3C3)(7C1) = 70 ways