Difference between revisions of "2010 AMC 10B Problems/Problem 22"

(Solution 2)
(Solution 2)
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== Solution 2 ==
 
== Solution 2 ==
We can use to our advantage the answer choices AMC has given us, and eliminate the obvious wrong answers.
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We can use to our advantage the answer choices <math>\text{AMC}</math> has given us, and eliminate the obvious wrong answers.
  
 
We can first figure out how many ways there are to take two candies from seven distinct candies to place them into the red/blue bags: <math>7\cdot 6=42</math>.
 
We can first figure out how many ways there are to take two candies from seven distinct candies to place them into the red/blue bags: <math>7\cdot 6=42</math>.

Revision as of 22:18, 13 August 2019

Problem

Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?

$\textbf{(A)}\ 1930 \qquad \textbf{(B)}\ 1931 \qquad \textbf{(C)}\ 1932 \qquad \textbf{(D)}\ 1933 \qquad \textbf{(E)}\ 1934$

Solution 1

We can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.

Each candy has three choices; it can go in any of the three bags.

Since there are seven candies, that makes the total distributions $3^7=2187$


To find the overcount, we calculate the number of invalid distributions: the red or blue bag is empty.

The number of distributions such that the red bag is empty is equal to $2^7$, since it's equivalent to distributing the $7$ candies into $2$ bags.

We know that the number of distributions with the blue bag is empty will be the same number because of the symmetry, so it's also $2^7$.

The case where both the red and the blue bags are empty (all $7$ candies are in the white bag) are included in both of the above calculations, and this case has only $1$ distribution.

The total overcount is $2^7+2^7-1=2^8-1$


The final answer will be $\text{total}-\text{overcount}=2187-(2^8-1) = 2187-256+1=1931+1=\boxed{\textbf{(C)}\ 1932}$

Solution 2

We can use to our advantage the answer choices $\text{AMC}$ has given us, and eliminate the obvious wrong answers.

We can first figure out how many ways there are to take two candies from seven distinct candies to place them into the red/blue bags: $7\cdot 6=42$.

Now we can look at the answer choices to find out which ones are divisible by $42$, since the total number of combinations must be $42$ multiplied by some other number.

Since answers A, B, D, and E are not divisible by 3 (divisor of 42), the answer must be $\boxed{\textbf{(C)}\ 1932}$.

See also

2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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