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Hi all,

By using 6 choose 2, aren't we ignoring the possibility of having a badge with two of the same color; i.e., RR,OO, YY, etc? In this case, we have to add 6 for this possibility, and then add 5 for the extra subjects that couldn't be included. I think the OA should be 26.

Can anyone give some insight?
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Hi all,

By using 6 choose 2, aren't we ignoring the possibility of having a badge with two of the same color; i.e., RR,OO, YY, etc? In this case, we have to add 6 for this possibility, and then add 5 for the extra subjects that couldn't be included. I think the OA should be 26.

Can anyone give some insight?

A scientist used a unique two-color code... I think from this we can assume that the two colors must be different.
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I guess that makes sense, but I as a reader would interpret that to mean that they used two 'slots' for the colors.

I imagine that by not using this possibility they're being highly inefficient - no wonder they weren't able to account for all of their test subjects.
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When you have this small selection of choices it is also easy to write them down in a table and cross out the ones you have used:
A_B_C_D_E_F--> the 6 colours, and we add them again beneath each colour to make the combinations.
A_A_A_A_A__A
B_B_B_B_B__B
C_C_C_C_C_C
D_D_D_D_D_D
E_E_E_E_E__E
F_F_F_F_F___F

So, you start by crossing out the same colours, e.g AA,BB,CC etc, and then the reverse colours, e.g, cross out BA because you have AB.
Then starting with A you've got:
AB - AC - AD - AE - AF
BC_BD_BE_BF
CD_CE_CF
DE_DF
EF
Nothing from F because all the colours have been used before.

So, 15 combinations, which are used for 15 people, plus 5 missing: 15+5=20.
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That in case 6!/(2!*4!) doesn't cross your mind... or you are slow with calculations, as I am.
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mikemcgarry
goodyear2013
A scientist used a unique two-color code to identify each of the test subjects involved in a certain study. If the scientist found that choosing from among six colors produced enough color codes to identify all but 5 of the test subjects, how many test subjects were in the study? (Assume that the order of the colors in the codes does not matter.)

7
10
15
17
20

OE
If 6 colors have to be divided into groups of 2, the number of unique groupings 6C2 = 6! / (2!)(4!) = 15.
Number of combinations was sufficient to account for all but 5 of the subjects
Number of subjects = 15 + 5 = 20.

Hi, I found this question stem not clear. Can anyone explain this for me, please.
Dear goodyear2013,
I'm happy to help. :-) I'm not entirely sure I understand what you found unclear. I will try telling the story in my own words.

A scientist was conducting a study, and he had to test individual subjects. As a way to identify the subjects, the scientist gave each subject something, say a badge, with two colors on it, and the scientist wanted the color combination to be different for each subject. Say there were N subjects in total. The scientist used six different individual colors, and different pairs formed from these six colors formed enough combinations to make unique color combinations for (N - 5) of the subject, so the scientist probably had to "double up" or do something different for those last 5 subjects.

This question is about counting techniques. See:
https://magoosh.com/gmat/2012/gmat-quant-how-to-count/
https://magoosh.com/gmat/2012/gmat-permu ... binations/
If we have six colors, the number of unique pairs of color we can create is
6C2 = 15
Each of the first 15 subjects got her or his unique pair of colors, leaving the last 5 with some other arrangement. Total = 15 + 5 = 20

Does all this make sense?
Mike :-)
Hi Mike,

I thought that since it is a code the order matters. let say we have Red/Green two color, Color code RG is different from GR and So 2! *6C2 + 5 = 35 .
I agree with mikemcgarry that the question stem is ambiguous.

thanks
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Hi Mike,

I thought that since it is a code the order matters. let say we have Red/Green two color, Color code RG is different from GR and So 2! *6C2 + 5 = 35 .
I agree with mikemcgarry that the question stem is ambiguous.

thanks
Dear Lucky2783,
I'm happy to reply. :-) Here's the question prompt again.
A scientist used a unique two-color code to identify each of the test subjects involved in a certain study. If the scientist found that choosing from among six colors produced enough color codes to identify all but 5 of the test subjects, how many test subjects were in the study? (Assume that the order of the colors in the codes does not matter.)
Notice that the question states, quite explicitly, that the order of the colors in the code doesn't matter. Remember never to let your own interpretation trump what is explicitly stated in the text of the problem.
Does this make sense?
Mike :-)
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goodyear2013
A scientist used a unique two-color code to identify each of the test subjects involved in a certain study. If the scientist found that choosing from among six colors produced enough color codes to identify all but 5 of the test subjects, how many test subjects were in the study? (Assume that the order of the colors in the codes does not matter.)

7
10
15
17
20

OE
If 6 colors have to be divided into groups of 2, the number of unique groupings 6C2 = 6! / (2!)(4!) = 15.
Number of combinations was sufficient to account for all but 5 of the subjects
Number of subjects = 15 + 5 = 20.

Hi, I found this question stem not clear. Can anyone explain this for me, please.

hi

6 * 6 = 36
means we can use the same color for a code, but it cannot be the case, for example, "Red + Red" cannot itself constitute any code

6 * 5 = 30
means we are distinguishing between "Red + Green" and "Green + Red ", but, according to the question, essentially they are the same

so, actually this is a combination as follows

6!
______
2! 3!

= 15

but, this amount falls short of encompassing all the studies, as we are left with 5 studies not coded
so the number of study is (15 + 5) = 20

thanks
:cool:
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