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

Latest revision as of 12:13, 20 December 2024

1 Problem
2 Solution 1 (Constructive Counting)
3 Solution 2 (Casework)
4 Solution 3 (Complementary Counting)
5 Video Solution by Math-X (First understand the problem!!!)
6 Video Solution (🚀Under 3 min🚀)
7 Video Solution by Robin "The Smartest Dude" Shang lol
8 Video Solution by OmegaLearn
9 Video Solution by SpreadTheMathLove
10 Video Solution by WhyMath
11 Video Solutions by The Learning Royal
12 Video Solution by Interstigation
13 Video Solution by STEMbreezy
14 See also

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 1 (Constructive Counting)

Firstly, observe that a single student can't 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 students receive $2$ awards and one student receives $1$ 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 be distributed. 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 $60+90=\boxed{\textbf{(B) }150}$ .

~Edits by WrenMath

Solution 2 (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 craps together, we get $60 + 90 = 150$ ways to distribute the awards, or choice $\boxed{\textbf{(B) }150}$ .

Solution 3 (Complementary Counting)

Without the restriction that each student receives at least one award, we could take each of the awards and choose one of the $3$ students to give it to. This would be $3^5$ ways to distribute the awards in total. Now we 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 $3$ choices for which student that is, so $2^5$ ways of choosing a student to receive each of the awards; in total, $3\cdot32=96$ .

However, if $2$ students both don't receive an award, then this case would be counted twice among the $96$ , so we need to add back in these cases. Said in other words, $2$ students not receiving an award is equivalent to $1$ student receiving $5$ awards, and there are $3$ choices for whom that student would be. To finish, the total number of ways to distribute the awards is $243 - 96+3$ , or $\boxed{\textbf{(B) }150}$ .

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=aB0A9Rb1KMIxOcpQ&t=5070

~Math-X

Video Solution (🚀Under 3 min🚀)

https://youtu.be/0CWA2KWvAXY

Video Solution by Robin "The Smartest Dude" Shang lol

https://youtu.be/3mMfy_evxl4

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

@@ Line 39: / Line 39: @@
 https://youtu.be/0CWA2KWvAXY
-==Video Solution by Robin "The Smartest Dude" Shang==
+==Video Solution by Robin "The Smartest Dude" Shang lol==
 https://youtu.be/3mMfy_evxl4
-Please like and subscribe!
 == Video Solution by OmegaLearn ==

2020 AMC 8 (Problems • Answer Key • Resources)
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