A child throws six differently colored candies up in the air. How many

Question

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

GMATinsight · Accepted Answer

Total Candies = 6

When  first  candy is thrown in air then the possible outcomes = 2 (Either child catches in mouth or not)

When  Second  candy is thrown in air then the possible outcomes = 2 (Either child catches in mouth or not)

When  Third  candy is thrown in air then the possible outcomes = 2 (Either child catches in mouth or not)

When  Forth  candy is thrown in air then the possible outcomes = 2 (Either child catches in mouth or not)

When  Fifth  candy is thrown in air then the possible outcomes = 2 (Either child catches in mouth or not)

When  Sixth  candy is thrown in air then the possible outcomes = 2 (Either child catches in mouth or not)

Total Possible outcomes = 2x2x2x2x2x2 = 64

But these 64 outcomes include one case in which all the candies are missed from being caught in mouth so we need to remove it

Total Favourable cases = 64-1 = 63

Answer: Option D

InfiniteLoop80 · Answer

The child catches: 1 candy, 2 candies, 3 candies, 4 candies, 5 candies or all 6 candies in her mouth.
This is a combination problem 
 She catches 1 candy = 6C1 different ways = 6 different ways
She catches 2 candies = 6C2 different ways
She catches 3, 4 , 5 or 6 = 6C3, 6C4, 6C5, 6C6 different ways.
Total possible ways = 6C1 + 6C2 + 6C3 + 6C4 + 6C5 +6C6 
either solve all these one by one and add 
OR
we know nC0 + nC1 + ...nCn = 2^n, 
 therefore nC1 +... nCn = 2^6 - 1 = 63 ways   choice D

Heseraj · Answer

The answer is a combination of all possible outcomes: one candy, two candies, three candies, four candies, five candies and six candies.

One Candy: 6C1 = 6

Two Candies: 6C2=  \frac{6*5}{1*2}=15

Three Candies: 6C3=  \frac{6*5*4}{1*2*3}=20

Four Candies: 6C4=  \frac{6*5*4*3}{1*2*3*4}=15

Five Candies: 6C5=  \frac{6*5*4*3*2}{1*2*3*4*5}=6

Six Candies: 6C6=  \frac{6*5*4*3*2*1}{1*2*3*4*5*6}=1

Total: 6+15+20+15+6+1=63

The correction option is D

praveenmittal95605 · Answer

case 1 = she catches 1 candy in her mouth = number of ways = 6C1 = 6
case 2 = number of ways of catching 2 candies in her mouth = 6C2 = 15
case 3= number of ways of catching 3 candies in her mouth = 6C3 = 20
case 4 = number of ways of catching 4 candies in her mouth = 6C4 = 15
case 5 = number of ways of catching 5 candies in her mouth = 6C5 = 6
case 6 = number of ways of catching all 6 candies in her mouth = 6C6 = 1

total = 63 Hence OPTION D

ScottTargetTestPrep · Answer

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.

Alternative 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

Jk8269 · Answer

Answer is D. It is the binomial expansion of 2^6. Since atleast one candy should fall in the mouth, we should exclude the one way in which no candies fall in her mouth. so naswer is 64-1 = 63

VyshakhR1995 · Answer

All cases - None will give answer...
Number of combinations = 64 
-----> 64-1 = 63
D

Jk8269 · Answer

Answer is D. It is the binomial expansion of 2^6. Since atleast one candy should fall in the mouth, we should exclude the one way in which no candies fall in her mouth. so naswer is 64-1 = 63 Please give Kudos f you like my Post

AliciaSierra · Answer

Total possible ways= 
6C1+6C2+6C3+6C4+6C5+6C6

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A child throws six differently colored candies up in the air. How many

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