Difference between revisions of "2020 AMC 8 Problems/Problem 23"

Revision as of 09:28, 28 January 2022

Problem

Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?

$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$

Solution (someone please rephrase this to make more sense)

Let the rows be $A, B, C$ , from left to right, and the columns be $1, 2, 3$ , from top to bottom. Notice that the line of three $\triangle$ s can be in either $A, B, C, 1, 2,$ or $3$ to allow for another line of three $\circ$ s. In each of those cases, there are two choices for where that line could go (since it has to be oriented the same and in a different row/column), and $2^3 = 8$ choices (triangle or circle for three remaining grid spaces) for the other row/column. However, notice that we overcount the cases in which the chosen row/column in which the line of triangles or circles is identical to the other column in the case that it was, too, all triangles or circles. For this, we subtract $\binom{3}{2} * 2 * 2 = 12$ (for rows and columns) cases. Therefore, our answer is $6 \cdot 2 \cdot 8 - 12 = \boxed{\text{(D) } 84}$ .

Solution 2 (Counting)

Firstly, observe that it is not possible for a single student to receive $4$ or $5$ awards because this would mean that one of the other students receives no awards. Thus, each student must receive either $1$ , $2$ , or $3$ awards. If a student receives $3$ awards, then the other two students must each receive $1$ award; if a student receives $2$ awards, then another student must also receive $2$ awards and the remaining student must receive $1$ award. We consider each of these two cases in turn. If a student receives three awards, there are $3$ ways to choose which student this is, and $\binom{5}{3}$ ways to give that student $3$ out of the $5$ awards. Next, there are $2$ students left and $2$ awards to give out, with each student getting one award. There are clearly just $2$ ways to distribute these two awards out, giving $3\cdot\binom{5}{3}\cdot 2=60$ ways to distribute the awards in this case.

In the other case, two student receives $2$ awards and one student recieves $1$ award . We know there are 3 choices for which student gets 1 award. There are $\binom{3}{1}$ ways to do this. Then, there are $\binom{5}{2}$ ways to give the first student his two awards, leaving $3$ awards yet to distribute. There are then $\binom{3}{2}$ ways to give the second student his $2$ awards. Finally, there is only $1$ student and $1$ award left, so there is only $1$ way to distribute this award. This results in $\binom{5}{2}\cdot\binom{3}{2}\cdot 1\cdot 3 =90$ ways to distribute the awards in this case. Adding the results of these two cases, we get $60+90=\boxed{\textbf{(B) }150}$ .

Video Solution by WhyMath

https://youtu.be/HkFQe7ZxBb4

~WhyMath

Video Solutions

https://youtu.be/tDChKU0pVN4
https://youtu.be/RUg6QfV5yg4

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=1443

~Interstigation

@@ Line 4: / Line 4: @@
 <math>\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240</math>
-==Solution 1 (Complementary Counting)==
+==Solution (someone please rephrase this to make more sense)==
-Without the restriction that each student receives at least one award, we could simply take each of the <math>5</math> awards and choose one of the <math>3</math> students to give it to, so that there would be <math>3^5=243</math> ways to distribute the awards. We now need to subtract the cases where at least one student doesn't receive an award. If a student doesn't receive an award, there are <math>3</math> choices for which student that is, then <math>2^5 = 32</math> ways of choosing a student to receive each of the awards, for a total of <math>3 \cdot 32 = 96</math>. However, if <math>2</math> students both don't receive an award, then such a case would be counted twice among our <math>96</math>, so we need to add back in these cases. Of course, <math>2</math> students both not receiving an award is equivalent to only <math>1</math> student receiving all <math>5</math> awards, so there are simply <math>3</math> choices for which student that would be. It follows that the total number of ways of distributing the awards is <math>243-96+3=\boxed{\textbf{(B) }150}</math>.
+Let the rows be <math>A, B, C</math>, from left to right, and the columns be <math>1, 2, 3</math>, from top to bottom. Notice that the line of three <math>\triangle</math>s can be in either <math>A, B, C, 1, 2,</math> or <math>3</math> to allow for another line of three <math>\circ</math>s. In each of those cases, there are two choices for where that line could go (since it has to be oriented the same and in a different row/column), and <math>2^3 = 8</math> choices (triangle or circle for three remaining grid spaces) for the other row/column. However, notice that we overcount the cases in which the chosen row/column in which the line of triangles or circles is identical to the other column in the case that it was, too, all triangles or circles. For this, we subtract <math>\binom{3}{2} * 2 * 2 = 12</math> (for rows and columns) cases. Therefore, our answer is <math>6 \cdot 2 \cdot 8 - 12 = \boxed{\text{(D) } 84}</math>.
 ==Solution 2 (Counting)==

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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