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

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

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

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

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
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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

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

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

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


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

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