Sticky pads

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piyush_nitt: Master | Next Rank: 500 Posts; Posts: 424; Joined: Sun Dec 07, 2008 5:15 pm; Location: Sydney; Thanked: 12 times

by piyush_nitt » Tue May 05, 2009 2:15 pm

2 sizes of sticky pads. Each has 4 colors – Blue, Green, Yellow, and Purple. The pads are packed in packages that contain either 3 notepads of same size and same color or 3 notepads of same size and of 3 different colors. How many different packages of the types described are possible?
a. 6
b. 8
c. 16
d. 24
e. 32

IMO: C

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Source: Beat The GMAT — Problem Solving | Tue May 05, 2009 2:15 pm

Jose Ferreira: GMAT Instructor; Posts: 99; Joined: Thu Mar 05, 2009 5:14 pmFollowed by: 9 members; Location: NYC; Thanked: 43 times; GMAT Score:800

by Jose Ferreira » Tue May 05, 2009 3:39 pm

I agree with your choice of answer C

First package type = (2 sizes)(4 different colors) = 8 total
or
big blue, big green, big yellow, big purple, small blue, small green, small yellow, and small purple.

Second package type = (2 sizes)(4 different combinations) = 8 total
or
big blue green yellow, big green yellow purple, big yellow purple blue, big purple blue green, and small of each of those.

8 + 8 = 16.

Bear in mind, this is assuming that there is no difference between having blue-green-yellow, for example, and yellow-green-blue. The question does not seem to specify that the order of the different colors matters.

Jose Ferreira
Founder and CEO, Knewton, Inc.
https://www.knewton.com/gmat

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uptowngirl92: Master | Next Rank: 500 Posts; Posts: 447; Joined: Sun Apr 19, 2009 9:08 pm; Location: Kolkata,India; Thanked: 7 times; GMAT Score:670

by uptowngirl92 » Wed Nov 04, 2009 7:26 pm

completly confused with so much of info..

Ok so lets say two sizes: small and large:
Each has B,G,Y,P colors.

We need case1 + case2
case1:same size same color
case2:same size diff. color

case1:
same size:Out of 2 we select 1: 2C1
same color: 4 ways either all B,all G,all Y..etc : 4
therefore case 1 gives us 8.

Case2:
same size:2C1
diff. color..hmmmm

Numerator:there are three slots,in first slot all 4 coors poss,in second slot 3 colors poss,in third slot 2 colors poss..therefore: 4!

Denominator:4 x 4 x 4

ok.ugly result..Help!Where am i gong wrong?

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antondesh: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Tue Aug 18, 2009 9:46 am; Thanked: 4 times

by antondesh » Thu Nov 05, 2009 1:03 pm

I agree with C.

There are two sizes and 4 colors possible for each of the sizes.

Let's take care of the size portion first. That's quite easy: if there are two sizes possible and each size can be one of the four colors, our total is 8.

The second part is slightly harder. Let's look at the smaller size. I used the combination formula:

4 possible colors and we need to pick out 3

4!/(3!*(4-3)!) = 4

The larger size would be exactly the same, i.e. 4.

Thus, 8 + 4 + 4 = 16.

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