Problem Solving

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?

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.

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?

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.)

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

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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