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Re: If a quality control check is made inspecting a sample of 2 [#permalink]
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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]
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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]
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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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Re: If a quality control check is made inspecting a sample of 2 [#permalink]
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]
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]
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]
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]
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