Combinations Problem

Expert replies

sathishkumarjva9888: Senior | Next Rank: 100 Posts; Posts: 46; Joined: Sun Dec 18, 2011 7:11 pm; Thanked: 1 times

by sathishkumarjva9888 » Mon Apr 30, 2012 10:14 am

OA: C

I know the question asks me to use combinations concept. But the answer which i calculated is not even among the listed answer choices.

I couldnt understand whether i am missing any restrictions on this selection? Please help.

Like this post

Source: Beat The GMAT — Problem Solving | Mon Apr 30, 2012 10:14 am

aneesh.kg: Master | Next Rank: 500 Posts; Posts: 385; Joined: Mon Apr 16, 2012 8:40 amFollowed by: 29 members; Location: Pune, India; Thanked: 186 times

by aneesh.kg » Mon Apr 30, 2012 10:26 am

If B,G,Y,P and b,g,y,p are the notepads (where the capital letters denote the larger-sized notepad), then (B,B,B), (G,G,G),...(y,y,y) and (p,p,p) will be the 8 ways in which we have 3 notepads of same size and same colour.
Thus, the Number of ways in which '3 notepads of same size and same colour'= 8

Number of ways in which '3 notepads of same size and different colours' can be obtained by selecting a size first from the 2 sizes (2C1) and then selecting any one of the four colours (4C3). So, total number of such ways = 2C1*4C3 = 8

Total number of ways = 8 + 8 = 16

Last edited by aneesh.kg on Mon Apr 30, 2012 10:40 am, edited 2 times in total.

Aneesh Bangia
GMAT Math Coach
[email protected]

GMATPad:
Facebook Page: https://www.facebook.com/GMATPad

Like this post

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Mon Apr 30, 2012 10:27 am

Hi!

Let's start with Step 1 of the Kaplan Method for PS: analyzing the question stem.

We're told that there are 2 different sizes of pad - let's call them small and large ("s" and "l"). There are also 4 different colours, b, g, y and p.

We're allowed to make 2 different types of package: 3 pads of the same size/colour and 3 pads of the same size/diff colours.

Finally, we're told that order of colour is irrelevant.

On to Step 2 of the Method: Identify the exact Q. We want to know the total number of possible packages that can be made.

Since there are 2 types of package, let's consider both cases.

1) all 3 same colour/size.

We have 2 different sizes and 4 different colours, so that gives us 2*4 = 8 possible uni-coloured packages (i.e. sb, lb, sg, lg, sp, lp, sy, ly).

2) all 3 same size/diff colours.

We still have 2 different sizes, we just need to figure out the possible combinations of 3 different colours. We could use the combinations formula (4C3, since we've got 4 options and we're choosing 3 of them) or we could just use logic or brute force.

Logic: if I'm using 3/4 colours, then I'm leaving out 1 colour. There are 4 different colours I could leave out, so there must be 4 possibilities.

Brute force: bgp, bgy, byp, gpy... 4 options.

2 sizes * 4 colour options = 8 multi-coloured packages.

Case 1 gave us 8 possibilities, as did Case 2. Accordingly, there are 8+8=16 total possible packages.. choose C!

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course

Like this post

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Mon Apr 30, 2012 10:29 am

Hi!

You've ignored the "of the same size" requirement for each possible package type, which is why you ended up with twice the possibilities.

aneesh.kg wrote:I will list three methods of solving this problem in my order of preference.

Method 1:
Select 3 colours out of 4 colours (4C3) and then select one size from each of those colours (2C1 three times).
Total number of ways: 4C3 * 2C1 * 2C1 * 2C1 = 32

Method 2:
Select any three notepads from the 8 notepads (8C3) and subtract those ways in which there is a pair of notepads with same colour (4C1*6C1)
Total number of ways: 8C3 - (4C1)*(6C1) = 32

Method 3:
Select any one of the 8 notepads (8C1) and then select any one of the remaining 6 notepads which are not of the previous colour (6C1) and then select another notepad which is not of the previous two colours (4C1). Then, since we have arranged instead of selecting, we must divide the total number of ways to nullify the arrangement.
Total number of ways = [(8C1)*(6C1)*(4C1)]/(3!) = 32

Pick your favourite.

Like this post

aneesh.kg: Master | Next Rank: 500 Posts; Posts: 385; Joined: Mon Apr 16, 2012 8:40 amFollowed by: 29 members; Location: Pune, India; Thanked: 186 times

by aneesh.kg » Mon Apr 30, 2012 10:43 am

Hi Stuart.
Didn't read the question at all. Just corrected it.
Thanks.

Aneesh Bangia
GMAT Math Coach
[email protected]

GMATPad:
Facebook Page: https://www.facebook.com/GMATPad

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review