A child throws six differently colored candies up in the air

[BTGmoderatorDC]

A child throws six differently colored candies up in the air. How many different possible groups of at least one candy are there that she could catch in her mouth?

A. 50
B. 51
C. 62
D. 63
E. 72

Can some experts figure out the best solution in this?

OA D

[elias.latour.apex]

This is a simple counting problem and, like most of these types of problems on the GMAT, it is designed to be a time waster. There are several possible solutions to this problem.

The first thing we could do is skip it entirely. If we choose to do so, we should pick either option B or D, because these are the options that cater to people who forget to include the 1 extra case of all 6 falling into her mouth. One of those two choices, which is 1 more than the immediately preceding answer, must be the correct one. This could give you a 50-50 shot at question credit for the investment of as little as 10-15 seconds. That's quite a bargain.

But let's suppose that it's the last question on the test and that any time you save could not be invested into other problems. Very well. Let's start counting.

There are 6 ways you can catch 1 candy in your mouth. Additionally, there are 6 ways you can catch 5 candies in your mouth. It's the same number because catching 5 out of 6 is the same as excluding one of the candies. So we're up to 12.

How many ways can you catch 2 candies in your mouth? Well, we can catch any of the 6 candies first, and then we will have our choice of 5 candies to catch next. So you might think that the answer is 30, but wait -- if you catch first the blue candy and then the red, that is no different from catching first the red candy and then the blue. So we must divide that number in half to get 15. As we reasoned above, the number of ways you can catch 4 candies is the same as the number of ways you can catch 2 candies. Whether you select 4 candies to catch OR select 2 candies NOT to catch, you are selecting the same group. So we are up to 30+12 = 42.

Now we must calculate the number of ways you can catch 3 candies in your mouth. We have 6 options of candies to catch first, then 5 for the second, and then 4 for the third candy. However, catching first red, then blue, then green is no different from green, blue, then red or even blue, red, and green. We have several duplicates to eliminate. The number of duplicates is 3! (3x2x1).

(6*5*4) / (3*2*1) reduces to 2*5*2 = 20. Adding this to our previous number of 42 gives us 62.

This is the moment of truth. Some people will have hacked their way through the test enough to reach this number. They will then select C. This is wrong. Do not forget (as we mentioned at the outset) that you must also include the one case of catching all 6. The best answer is D.

[GMATGuruNY]

A child throws six differently colored candies up in the air. How many different possible groups of at least one candy are there that she could catch in her mouth?

A. 50
B. 51
C. 62
D. 63
E. 72

Let C = caught and N = not caught.

Since each candy must be caught or not caught, there are TWO OPTIONS for each candy:
C or N.
Number of options for the 1st candy = 2. (C or N)
Number of options for the 2nd candy = 2. (C or N)
Number of options for the 3rd candy = 2. (C or N)
Number of options for the 4th candy = 2. (C or N)
Number of options for the 5th candy = 2. (C or N)
Number of options for the 6th candy = 2. (C or N)
To combine the options for each candy, we multiply:
2*2*2*2*2*2 = 64.

But the prompt requires that at least one candy be caught.
Of the 64 ways above, one way is invalid:
NNNNNN, which implies that NONE of the candies is caught.
Subtracting the one invalid way from the 64 possible ways, we get:
64-1 = 63.

The correct answer is D.

[Scott@TargetTestPrep]

A child throws six differently colored candies up in the air. How many different possible groups of at least one candy are there that she could catch in her mouth?

A. 50
B. 51
C. 62
D. 63
E. 72

Since the order of the candies within a group of candies doesn't matter, this is a combination problem.

Number of ways the group of candies consists of exactly 1 candy: 6C1 = 6

Number of ways the group of candies consists of exactly 2 candies: 6C2 = (6 x 5)/2 = 15

Number of ways the group of candies consists of exactly 3 candies: 6C3 = (6 x 5 x 4)/(3 x 2) = 20

Number of ways the group of candies consists of exactly 4 candies: 6C4 = 6C2 = 15

Number of ways the group of candies consists of exactly 5 candies: 6C5 = 6C1 = 6

Number of ways the group of candies consists of all 6 candies: 6C6 = 1

Therefore, the number of ways the group of candies consists of at least 1 candy is 6 + 15 + 20 + 15 + 6 + 1 = 63.

Alternate solution:

This is an "at least one" problem. Therefore, the number of ways the group of candies consists of at least 1 candy is the number of ways the group of candies consists of any number (from 0 to 6, inclusive) of candies, minus the number of ways the group of candies consists of exactly 0 candies.

The number of ways the group of candies consists of any number of candies is 2^6 = 64, and the number of ways the group of candies consists of exactly 0 candies is 6C0 = 1. Thus, the number of ways the group of candies consists of at least 1 candy is 64 - 1 = 63.

Answer: D