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

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!

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.

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