Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution (🚀Under 3 min🚀))
(→‎Solution 1 (Principle of Inclusion-Exclusion))
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>
 
<math>\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240</math>
  
−
==Solution 1 (Principle of Inclusion-Exclusion)==
+
==
−
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. Therefore, the total number of ways of distributing the awards is <math>243-96+3=\boxed{\textbf{(B) }150}</math>.
 
  
 
==Solution 2 (Constructive Counting)==
 
==Solution 2 (Constructive Counting)==

Revision as of 19:50, 12 November 2023

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 2 (Constructive 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 $60+90=\boxed{\textbf{(B) }150}$.

Solution 3 (Casework)

Upon inspection (specified in the above solution), there are two cases of the distribution of awards to the students: one student gets 3 awards and the other each get 1 award or one student gets 1 award and the other two get 2 awards.


In the first case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 3 awards. From here, there are $\binom{5}{3} = 10$ ways to choose the 3 awards from the 5 total awards. Now, one person has $2$ choices for awards and the other has $1$ choice for the award. Thus, the total number of ways to choose awards in this case is $3 \cdot 10 \cdot 2 \cdot 1 = 60$.


In the other case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 1 award, and $5$ choices for his/her award. Then, one person has $\binom{4}{2} = 6$ ways to have his/her awards and the other person has $\dbinom{2}{2} = 1$ ways to have his/her awards. This gives $3 \cdot 5 \cdot 6 \cdot 1 = 90$ ways for this case.

Adding these cases together, we get $60 + 90 = 150$ ways to distribute the awards, or choice $\boxed{\textbf{(B) }150}$.

~TaeKim


Video Solution (🚀Under 3 min🚀)

https://youtu.be/0CWA2KWvAXY

Video Solution

https://youtu.be/3mMfy_evxl4

Please like and subscribe!

Video Solution by OmegaLearn

https://youtu.be/dFFjlxm43b0?t=899

~ pi_is_3.14

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=Dg_2wXNY3tE

Video Solution by WhyMath

https://youtu.be/HkFQe7ZxBb4

~WhyMath

Video Solutions by The Learning Royal

https://youtu.be/tDChKU0pVN4

https://youtu.be/RUg6QfV5yg4

Video Solution by Interstigation

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

~Interstigation

Video Solution by STEMbreezy

https://youtu.be/wq8EUCe5oQU?t=243

~STEMbreezy

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_23&oldid=202811"