In Finn's preschool class, each student is assigned a unique

Question

In Finn's preschool class, each student is assigned a unique color palette of 3 colors for a finger-painting project. Different students may have 1 or 2 colors in common, but no 2 students have the same 3 colors. If there are 10 students in the class, how many different colors are required?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

Please explain how you arrived your answer.

Sam Kana · Answer

if 5 is the answer..

I guess here they r just checking ur combinatorial skills...

if u have 3 colors u can select 3 colors in only one way 3c3=1
and for 4: 4c3 = 4
and for 5c3 = 10..

or other option is nc3=10, and solve for 3...

pawan82 · Answer

I have a different answer.

Lets assume students there are four colors Red (R) Blue (B) Green (G) Yellow (Y) So no. of ways they can be split is 1 RBG 2 RRB 3 RGR 4 RBB 5 RGG 6 GGB 7 BBG 8 YYR 9 YRB 10 YGR Hence the ans should be A i.e. 4

Sumithra · Answer

I think, the question says 3 colors for each student. Don't RRB, RBB,... have 2 colors?

ywilfred · Answer

Trial and error:

Try 4C3 --> 4 
Try 5C3 --> 10

So answer should be B

andrewnorway · Answer

In Finn's preschool class, each student is assigned a unique color palette of 3 colors for a finger-painting project. Different students may have 1 or 2 colors in common, but no 2 students have the same 3 colors. If there are 10 students in the class, how many different colors are required? 

(A) 4 
(B) 5 
(C) 6 
(D) 7 
(E) 8 

Please explain how you arrived your answer.
------------------------------------------------------------------------------------
You have 3 per group and need 10 possible combinations. You can set the forumla up as follows. For n colors with r per group you have:
n(n-1)(n-2)...(n-r+1) / r! 

You need that to be greater than 10.

If you set up the forumal with all the numbers and start multiplying from right to left, you see that 4 is not enough and 5 is more than enough, so the answer is 5.

8X7X6X5X4X3X2X1 / 3X2X1 = 10

If you reduce you get:

8X7X6X5X4 = 10

You can then quickly go from right to left and see that 4 colors is not enough but 5 is.

MBAlad · Answer

Andre,

In your working you are neglecting that you have to account for repeated combinations within the choices so it is not simply n!/r! because r! will contain the number of permutations rather than unique combinations.

eg if 5 colours then number of permutations if 3 are chosen is 5!/3! = 20
so if colours used are Red, Green, Yellow, Blue you would be counting unique combinations twice ie RGY, RYG etc

Therefore we use 5C3 = 10

andrewnorway · Answer

Yeah, math error. I had the formula above, but kept 2X1 in my actual math. Should have been:

8X7X6X5X4X3 / 3X2X1 = 10

Instead of
8X7X6X5X4X3X2X1 / 3X2X1 = 10

I'm f$#ked if I keep doing stupid sh%t like that one the real deal

Pokhran II · Answer

Thanks everyone. I agree that one needs to backsolve by applying trial-and-error technique. 

The OA is B, which is 5.

In Finn's preschool class, each student is assigned a unique

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