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If a quality control check is made inspecting a sample of 2

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Matrix02
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If a quality control check is made inspecting a sample of 2 [#permalink] New post   Updated on: 13 Dec 2014, 05:48
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If a quality control check is made inspecting a sample of 2 light bulbs from a box of 12 lighbulbs, how many different samples can be chosen?

A) 6
B) 24
C) 36
D) 66
E) 72

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I understand picking 12 options for the 1st light bulb and 11 for the next light bulb but why do you divide this number in half (and how do you know to divide this in half and this will gurantee that all the samples are different?) Thanks
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D

Originally posted by Matrix02 on 05 Nov 2006, 17:59.
Last edited by Bunuel on 13 Dec 2014, 05:48, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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wshaffer
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Re: If a quality control check is made inspecting a sample of 2 [#permalink] New post   05 Nov 2006, 18:32
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Try using the formula for combinations:

n!/r!(n-r)!

12!/2!(12-2)!

12!/2!*10!

12*11/2*1

=66
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rlitagmatstudy
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Re: If a quality control check is made inspecting a sample of 2 [#permalink] New post   23 May 2016, 21:50
Is there a different way of solving this problem without combinations? Can it be solved with test the answers?
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Re: If a quality control check is made inspecting a sample of 2 [#permalink] New post   08 Jun 2017, 11:11
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rlitagmatstudy wrote:
Is there a different way of solving this problem without combinations? Can it be solved with test the answers?




https://www.khanacademy.org/math/precal ... mbinations

go for it
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Re: If a quality control check is made inspecting a sample of 2 [#permalink] New post   26 Jun 2018, 18:16
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Matrix02 wrote:
If a quality control check is made inspecting a sample of 2 light bulbs from a box of 12 lighbulbs, how many different samples can be chosen?

A) 6
B) 24
C) 36
D) 66
E) 72


The number of samples that can be chosen is:

12C2 = 12!/(2! x 10!) = (12 x 11)/2! = 66

Answer: D
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Re: If a quality control check is made inspecting a sample of 2 [#permalink] New post   27 Jun 2020, 12:20
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Just solved this correctly using a method I learned from a thursdays with ron video ( I think it was from 2017).
Granted, it is not timely, but it worked.

1. List bulbs out from option A - L (1-12)
2. List options that occur once
3. Sum to total

Ex:

A B
A C
A D
A E
A F
A G
A H
A I
A J
A K
A L

then
BC
BD
BE..BF.....BL

CD.....CL

You slowly get less and less because the combination already occurs, so you dont count it (eg., LC doesnt count, you already counted CL). Sums to 66.
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rahulgoel007
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Re: If a quality control check is made inspecting a sample of 2 [#permalink] New post   13 Feb 2021, 01:00
We are to choose 2 objects out of the 12 given. Since it is selection, nCr = n!/r!(n-r)!

Here, n=12 and r=2. Substituting, we have

12C2=12!/2!(12-2)!

=12!/2!*10!

=12*11/2*1

=66

So, the answer is D.
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Re: If a quality control check is made inspecting a sample of 2 [#permalink] New post   03 Mar 2021, 19:21
Where do you get the 11 from?
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Re: If a quality control check is made inspecting a sample of 2 [#permalink] New post   30 Mar 2021, 20:21
crashcarson wrote:
Where do you get the 11 from?


The combinatorial formula is

The total number of units factorial (in this case 12!) divided by (total units - number of choices which in this case is 12-10) factorial times the number of choices factorial. Therefore

12! = 12x11x10x9x8x7...
divided by
(12-2)! = 10! = 10x9x8x7... times 2!

the 10! cancels out all the values from 10 down in the numerator, hence you're left with 12 x 11 which is divided by 2! (which is 2 x 1) and gives you a value of 66

hope this helps
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Re: If a quality control check is made inspecting a sample of 2 [#permalink] New post   11 Jul 2021, 01:20
alk27 wrote:
Just solved this correctly using a method I learned from a thursdays with ron video ( I think it was from 2017).
Granted, it is not timely, but it worked.

1. List bulbs out from option A - L (1-12)
2. List options that occur once
3. Sum to total

Ex:

A B
A C
A D
A E
A F
A G
A H
A I
A J
A K
A L

then
BC
BD
BE..BF.....BL

CD.....CL

You slowly get less and less because the combination already occurs, so you dont count it (eg., LC doesnt count, you already counted CL). Sums to 66.


It would be much better to use 12C2 and get the answer quickly. But yes, the above method is the logic as to what is exactly going on in this question. I was about to post a similar solution until I saw this :)
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Re: If a quality control check is made inspecting a sample of 2 [#permalink] New post   12 Nov 2022, 10:40
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Re: If a quality control check is made inspecting a sample of 2 [#permalink]
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