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Help me solve this GMAT Quant Question (GMATPrep) #3

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by EricImasogie » Fri Dec 12, 2014 5:55 pm
This question is turning out to be mind numbing. Initially I felt that I could approach this using factorial rules of n!/ r!(n-r)! but I am having a hard time trying to figure out exactly what "r" or "n" is! GEESH!! Can somebody guide me in the right direction? or should I just pick C (which happened to be the right answer btw) and keep it moving to salvage time?


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by Brent@GMATPrepNow » Fri Dec 12, 2014 6:40 pm
A certain office supply store stocks 2 sizes of stick notepads,each in 4 colours : blue,green,yellow or pink.The store packs the note pads in packages that contain either 3 notepads of the same size and the same colour or 3 notepads of the same size and different colours.If the order in which the colours are packed doesnt matter,how many different packages of the types described above are possible?

a)6
b)8
c)16
d)24
e)32

Thanks
There are two different cases to consider:
1) All 3 pads the same color
2) The 3 pads are 3 different colors

Case 1: All 3 pads the same color
Take the task of packaging pads and break it into stages.

Stage 1: Select a size
There are 2 possible sizes, so we can complete stage 1 in 2 ways.

Stage 2: Select 1 color (to be applied to all 3 pads)
There are 4 possible colors from which to choose, so we can complete stage 2 in 4 ways.

By the Fundamental Counting Principle (FCP) we can complete the two stages in (2)(4) ways (= 8 ways)


Case 2: The 3 pads are 3 different colors
Take the task of packaging pads and break it into stages.

Stage 1: Select a size
There are 2 possible sizes, so we can complete stage 1 in 2 ways.

Stage 2: Select 3 different colors
There are 4 possible colors, and we must choose 3 of them.
Since the order of the selected colors does not matter, we can use combinations.
We can select 3 colors from 4 colors in 4C3 ways (4 ways), so we can complete stage 2 in 4 ways.

By the Fundamental Counting Principle (FCP) we can complete the two stages in (2)(4) ways (= 8 ways)


So, both cases can be completed in a total of 8 + 8 ways =[spoiler] 16 = C[/spoiler]

--------------------------

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Brent
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by EricImasogie » Fri Dec 12, 2014 6:52 pm
WOw! thanks Brent! I really didn't even see it this way, thanks for the guidance! this type of question is designed to frustrate you into using other methods that will lead you no where, but when you break it down like this, it made it a lot easier.
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by [email protected] » Fri Dec 12, 2014 9:35 pm
Hi EricImasogie,

The answer choices to this question are all relatively small, so if you really were stuck on this question, then you could "brute force" the answer by writing out all of the possibilities. I bet that you could do it relatively quickly.

I'm going to call the 4 colors: a, b, c and d
I'm going to call the size/color combinations: A, a, B, b, C, c, D and d

We're told that there are 2 different "packages" of notepads:

3 pads of the same size and color & 3 pads of the same size but DIFFERENT colors. Let's count them up:

3 pads of the same size/color:
AAA
BBB
CCC
DDD
aaa
bbb
ccc
ddd

8 possibilities here.

3 pads of the same size but DIFFERENT colors (note: order DOES NOT matter):
ABC
ABD
ACD
BCD
abc
abd
acd
bcd

8 possibilities here.

Total = 8 + 8 = 16

Final Answer: C

There will be at least a couple of questions on Test Day that can be "brute force-d" in this way. If you're not exactly sure how the math 'works' and the answers are relatively small, then try sketching out the possibilities.

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by ceilidh.erickson » Mon Dec 15, 2014 4:24 pm
EricImasogie wrote:This question is turning out to be mind numbing. Initially I felt that I could approach this using factorial rules of n!/ r!(n-r)! but I am having a hard time trying to figure out exactly what "r" or "n" is! GEESH!! Can somebody guide me in the right direction? or should I just pick C (which happened to be the right answer btw) and keep it moving to salvage time?
To answer your question more broadly... the n!/r!(n-r)! approach works only when you're picking out of the *same pool*, and *order doesn't matter*. For example, "how many ways are there to pick 3 notepads out of 5 different colored notepads?" Then we'd do 5!/3!2!

When you have a question that involves two different ways to pick (all one color or 3 different colors), you just have to list out the possibilities in an organized way.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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by Mathsbuddy » Tue Dec 16, 2014 5:39 am
Surely it is irrelevant that there are 3 pads each time?
Then it leaves pads of same/different colour or same/different size.
Surely this means that all combinations are valid?
Number of combinations = 2^4 = 16
Is this logic flawed?
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by [email protected] » Tue Dec 16, 2014 10:36 am
Hi Mathsbuddy,

In the first grouping (all pads are the same size and color), the number of pads does NOT matter - the number of outcomes is the same regardless. There would be any number of...

just A
just a
just B
just b
just C
just c
just D
just d

So 8 options will always exist.

However, in the second grouping (all the same size, but 3 DIFFERENT colors), the number of pads DOES matter.

In this question, with 3 colors, we have the following options:

ABC
ABD
ACD
BCD
abc
abd
acd
bcd

8 total options

IF we were only picking 2 colors though, the number of possibilities changes:

AB
AC
AD
BC
BD
CD
ab
ac
ad
bc
bd
cd

Here, we have 12 options....a different answer.

So to answer your question, the number of pads DOES matter.

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by Matt@VeritasPrep » Mon Dec 22, 2014 1:33 pm
Mathsbuddy wrote:Surely it is irrelevant that there are 3 pads each time?
Then it leaves pads of same/different colour or same/different size.
Surely this means that all combinations are valid?
Number of combinations = 2^4 = 16
Is this logic flawed?
Here's a mathy explanation. Suppose we have s sizes and c colors, of which we're choosing x colors. The number of valid combinations is thus sc + s*(c choose x). If I read you correctly, you're proposing that sc + s*(c choose x) = sᶜ. This happens to be true for s = 2, c = 4, x = 3, but it isn't true for many other integral values of s, c, and x.

In this case your approach works, since we're essentially choosing one of the colors to leave out. (Mathily, x = c - 1, which gives us sc + s*(c choose (c-1)) = sᶜ, or 2sc = sᶜ, for which s = 2 and c = 4 gives an identity.)
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by bschool3 » Wed Aug 26, 2015 1:38 pm
Another way to think about it:
You have 2 choices to make: choose color, choose size:
Case1: Same size, same color
# of ways to choose 1 color out of 2 = 2C1; # of ways to choose 1 size from 4 = 4C1
# of ways to choose same color AND same size = 2C1*4C1 = 8
Case2:Same size, different color
# of ways to choose 1 color out of 2 = 2C1; # of ways to choose 3 different sizes from 4 = 4C3
# of ways to choose same color AND different sizes = 2C1*4C3 = 8

Number of ways to have case 1 OR case 2 = (2C1*4C1) + (2C1*4C3) = 8 + 8 = 16
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by Max@Math Revolution » Thu Aug 27, 2015 3:25 am
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

A certain office supply store stocks 2 sizes of stick notepads,each in 4 colours : blue,green,yellow or pink.The store packs the note pads in packages that contain either 3 notepads of the same size and the same colour or 3 notepads of the same size and different colours.If the order in which the colours are packed doesnt matter,how many different packages of the types described above are possible?

a)6
b)8
c)16
d)24
e)32

==> in case of 3 notepads of the same size and the same color
SIZE 1: 4 cases for 4 colors.
SIZE 2: again 4 cases, since there are 4 colors. Therefore 4+4 = 8

in case of 3 notepads of the same size and different colors
SIZE 1: since we choose 3 from 4 different colors, 4C3=4
SIZE 2: again, since we choose 3 from 4 different colors, 4C3=4. therefore 4+4=8

Thus we have 8+8 = 16. Therefore C is the answer.


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