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help hotel buy defective TVs

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help hotel buy defective TVs

by ashish1354 » Sat Sep 20, 2008 4:07 am
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
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by loki.gmat » Sat Sep 20, 2008 4:42 am
well u r missing one more case i.e, when none of the TV sets selected r defective.

so the answer = 10c4 - 3c1*7c3 - 7c4
= 210 - 105 - 35 = 70.


Thanks!
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by cramya » Sat Sep 20, 2008 9:03 am
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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by dally_gmat » Sat Sep 20, 2008 2:30 pm
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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by pooja181 » Fri Jul 11, 2014 10:07 pm
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
My skype id- varun289verma

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by GMATinsight » Fri Jul 11, 2014 10:35 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?
2 defective - 7C2 x 3C2 = 21 x 3 = 63
3 defective - 7C1 x 3C3 = 7 x 1 = 7

Total Favorable cases = 63+7 = [spoiler]70[/spoiler]
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by GMATinsight » Fri Jul 11, 2014 10:40 pm
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
Hi Varun,

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.
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Most Comprehensive and Affordable Video Course 2000+ CONCEPT Videos and Video Solutions
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