counting methods

Fractal wrote:A shop sells white, red, and black t-shirts. how many combinations of 4 shirts can someone buy?

a) 6
b) 12
c) 15
d) 64
e) 81

There are 4 possibilities
1) All four of same color and this can be done in 3C1 = 3 ways.
2) Three of same color and one a different color. This can be done in 3C1 * 2 = 6 ways
3) Two of same color and two of different two colors. We can have this in 3C1 = 3 ways
4) Two of same color and the other two of some other same color. We can have this in 3C2 = 3 ways.

Hence, total number of possible combinations = 3+6+3+3 = 15.

Fractal wrote:A shop sells white, red, and black t-shirts. how many combinations of 4 shirts can someone buy?
a) 6
b) 12
c) 15
d) 64
e) 81

There's a quick way to set up this question so that we can readily see that the answer is 6C2 (=15)

Challenge: Can anyone explain why the answer will be 6C2?

Cheers,
Brent

Juggernaut_86 wrote:Anurag,

I am getting 15 as the answer just by listing out the possibilities-

1) All 4 of the same color - WWWW, RRRR, BBBB = 3

2) 3 of the same color - WWWR, WWWB, RRRW, RRRB, BBBW, and BBBR = 6

3) 2 of the same color - WWRR, WWBB, RRBB, WWBR, RRBW, and BBRW = 6

Total 15...Am I overestimating?

Thanks!

By the way, I love this solution!

Students often undervalue the utility of simply listing possible outcomes.
First, if the answer choices are small (as they are here), there's a good chance that you may be able to list all outcomes.
Second, while listing outcomes, students often find certain patterns that allow them to stop listing and employ a different (faster) technique to solve the question.

Cheers,
Brent

Fractal wrote:A shop sells white, red, and black t-shirts. how many combinations of 4 shirts can someone buy?
a) 6
b) 12
c) 15
d) 64
e) 81

Note: To make this solution easier to see with different colors, let's say that the three t-shirt colors are red, green and blue (the answer will still be the same)

Okay, here's the setup.

Let's say we have 6 blanks lined up in a row: _ _ _ _ _ _
Now if we choose 2 of those blanks and replace them with an O, the remaining 4 blanks are divided into 3 regions.
Example, _ O _ _ O _
Notice that we have one blank (region 1) then 2 blanks (region 2) and then 1 blank space (region 3).
We'll say that:
The number of blanks in region 1 will represent the number of red shirts.
The number of blanks in region 2 will represent the number of green shirts.
The number of blanks in region 3 will represent the number of blue shirts.

So, when we choose the second and fifth blanks, we get: _ O _ _ O _ , which means we have 1 red shirt, 2 greens and 1 blue.

When we select the second and fourth blanks, we get _ O _ O _ _, which means we have 1 red shirt, 1 green and 2 blues.

When we select the second and third blanks, we get _ O O _ _ _, which means we have 1 red shirt, 0 green and 3 blues. (here, there are zero blanks in region 2)

When we select the first and second blanks, we get O O _ _ _ _, which means we have 0 red shirts, 0 green and 4 blues.

When we select the fifth and sixth blanks, we get _ _ _ _ O O , which means we have 4 red shirts, 0 green and 0 blue.

Notice that each selection of 2 blanks results in a new combination of reds, greens and blues.

So, the question now becomes, "In how many ways can we choose 2 of the 6 blanks?" and the answer is 6C2

Cheers,
Brent

Fractal wrote: but i don't know any other gmat problems which need this formula....

Here's a question (and solution) that I posted a long time ago that requires a similar approach: https://www.beatthegmat.com/very-tricky- ... 25349.html

Cheers,
Brent

ashutoshkumar7 wrote:I can see the answer 15 is right, but as soon as I saw the question I did 3*3*3*3
as for the 1st shirt we have option of 3 and for 2nd we have 3 options and so on.

Your approach treats each selection position as different. for example, your approach assumes that buying one white shirt, and then 3 red shirts is different from buying 3 red shirts and then 1 white shirt. If these were considered different purchases, then your approach would be perfect.

That said, I think the question could be reworded to avoid any ambiguity.

Cheers,
Brent