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?
i think ways to purchase atleat 2 def tv sets= total no of ways to buy 4 tv sets (10c4)-no of ways to buy exactly 1 def tv set(3c1*7c3)
atleast 2 defective tv sets= 10c4-3c1*7c3=210-105=105
but the answer is wrong
can someone suggest what is my mistake
help hotel buy defective TVs
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When it says atleast2 either 2 or more could be defective
Total no of ways =
No of ways of 2 being defective + No of ways 3 can be defective(since there are only 3 defective sets)
= 7C2.3C2 + 7C1.3C3
= 63+7
=70
7C2.3C2 IS THE CASE WHERE 2 NON DEFECTIVE SETS IS PICKED FROM 7 AND 2 DEFECTIVE SETS IS PICKED FROM 3
7C1.3C3 IS THE CASE WHERE 1 NON DEFECTIVE SETS IS PICKED FROM 7 AND 3 DEFECTIVE SETS IS PICKED FROM 3
Hope this helps!
Total no of ways =
No of ways of 2 being defective + No of ways 3 can be defective(since there are only 3 defective sets)
= 7C2.3C2 + 7C1.3C3
= 63+7
=70
7C2.3C2 IS THE CASE WHERE 2 NON DEFECTIVE SETS IS PICKED FROM 7 AND 2 DEFECTIVE SETS IS PICKED FROM 3
7C1.3C3 IS THE CASE WHERE 1 NON DEFECTIVE SETS IS PICKED FROM 7 AND 3 DEFECTIVE SETS IS PICKED FROM 3
Hope this helps!
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Whats is OA?
cramya wrote:When it says atleast2 either 2 or more could be defective
Total no of ways =
No of ways of 2 being defective + No of ways 3 can be defective(since there are only 3 defective sets)
= 7C2.3C2 + 7C1.3C3
= 63+7
=70
7C2.3C2 IS THE CASE WHERE 2 NON DEFECTIVE SETS IS PICKED FROM 7 AND 2 DEFECTIVE SETS IS PICKED FROM 3
7C1.3C3 IS THE CASE WHERE 1 NON DEFECTIVE SETS IS PICKED FROM 7 AND 3 DEFECTIVE SETS IS PICKED FROM 3
Hope this helps!
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Can any body brief me where to choose. Combination. And where to use. Simple like 7.6.5.
Every time I miss for such type question
Every time I miss for such type question
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2 defective - 7C2 x 3C2 = 21 x 3 = 63A 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?
3 defective - 7C1 x 3C3 = 7 x 1 = 7
Total Favorable cases = 63+7 = [spoiler]70[/spoiler]
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Hi Varun,Pooja181 wrote:
Can any body brief me where to choose. Combination. And where to use. Simple like 7.6.5.
Every time I miss for such type question
7x6x5 - Such a method includes Selection as well as arrangement of 3 Objects out of a total of 7 objects
nCr - This operator is used only to select the r objects out of n objects
If you carefully check both the things then they can be related as
7 x 6 x 5 = 7C3 x 3!
I hope it explains your doubt. You can reach us for more such doubts.
"GMATinsight"Bhoopendra Singh & Sushma Jha
Most Comprehensive and Affordable Video Course 2000+ CONCEPT Videos and Video Solutions
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Contact for One-on-One FREE ONLINE DEMO Class Call/e-mail
Most Efficient and affordable One-On-One Private tutoring fee - US$40-50 per hour
Most Comprehensive and Affordable Video Course 2000+ CONCEPT Videos and Video Solutions
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