Hi. I barely missed this problem as I couldn't recall the formula.. I guess it falls under "Number Properties" Can you help me recall?
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Can you recall the formula for this?
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Hi Yumi2012,
This is an example of a Combination Formula question. After reading these types of prompts, you have to determine whether ORDER MATTERS or DOES NOT MATTER.
Since the question asks us to figure out the number of "groups of 2", then the order DOES NOT MATTER (Bulb A and Bulb B is the same group as Bulb B and Bulb A). This means that we need the Combination Formula:
N! / [K!(N-K)!]
N = 12
K = 2
12! / [2!(10)!] = 12(11)/2 = 66
GMAT assassins aren't born, they're made,
Rich
This is an example of a Combination Formula question. After reading these types of prompts, you have to determine whether ORDER MATTERS or DOES NOT MATTER.
Since the question asks us to figure out the number of "groups of 2", then the order DOES NOT MATTER (Bulb A and Bulb B is the same group as Bulb B and Bulb A). This means that we need the Combination Formula:
N! / [K!(N-K)!]
N = 12
K = 2
12! / [2!(10)!] = 12(11)/2 = 66
GMAT assassins aren't born, they're made,
Rich
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ganeshrkamath
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Number of ways to select r items out of n unique items = nCr = n! / ((n-r)!*r!)yumi2012 wrote:Hi. I barely missed this problem as I couldn't recall the formula.. I guess it falls under "Number Properties" Can you help me recall?
In this case:
r = 2
n = 12
12C2 = 12! / ((12-2)!*2!)
= 12! / (10! * 2!)
= (12*11)/2
= 6*11
= 66
Choose D
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Kelley School of Business (Class of 2016)
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GMATGuruNY
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Approach 1: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
Number of options for the 1st bulb = 12. (Any of the 12 bulbs.)
Number of options for the 2nd bulb = 11. (Any of the 11 remaining bulbs.)
To combine these options, we multiply:
12*11.
Since the ORDER of the bulbs doesn't matter -- AB are the same two bulbs as BA -- we divide by the number of ways to ARRANGE the 2 bulbs (2!):
(12*11)/(2*1) = 66.
The correct answer is D.
Approach 2:
Let the 12 bulbs be A-B-C-D-E-F-G-H-I-J-K-L.
Bulb A can be paired with any of the following:
B-C-D-E-F-G-H-I-J-K-L.
Total options = 11.
All of the pairs that include Bulb A have been counted.
If we exclude Bulb A, Bulb B can be paired with any of the following:
C-D-E-F-G-H-I-J-K-L.
Total options = 10.
All of the pairs that include Bulbs A and/or B have been counted.
If we exclude Bulbs A and B, Bulb C can be paired with any of the following:
D-E-F-G-H-I-J-K-L.
Total options = 9.
Note the PATTERN.
The results are descending consecutive integers, starting with 11.
Implication:
The total number of pairs is equal to the sum of the integers between 11 and 1, inclusive.
Thus:
Total possible pairs = 11+10+9+8+7+6+5+4+3+2+1 = 66.
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As a tutor, I don't simply teach you how I would approach problems.
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Brent@GMATPrepNow
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If anyone is interested, we have a free video on calculating combinations (like 12C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789ganeshrkamath wrote:Number of ways to select r items out of n unique items = nCr = n! / ((n-r)!*r!)yumi2012 wrote:Hi. I barely missed this problem as I couldn't recall the formula.. I guess it falls under "Number Properties" Can you help me recall?
In this case:
r = 2
n = 12
12C2 = 12! / ((12-2)!*2!)
= 12! / (10! * 2!)
= (12*11)/2
= 6*11
= 66
Choose D
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Cheers,
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Brent Hanneson - Creator of GMATPrepNow.com
