2020 AMC 8 Problems/Problem 23
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?
Solution 1
Firstly, observe that it is not possible for a single student to receive or awards because this would mean that one of the other students receives no awards. Thus, each student must receive either , , or awards. If a student receives awards, then the other two students must each receive award; if a student receives awards, then another student must also receive awards and the remaining student must receive award. We consider each of these two cases in turn. If a student receives three awards, there are ways to choose which student this is, and ways to give that student out of the awards. Next, there are students left and awards to give out, with each student getting one award. There are clearly just ways to distribute these two awards out, giving ways to distribute the awards in this case.
In the other case, a student receives awards. We first have to choose which of the two students we will select to give two awards each to. There are ways to do this, after which there are ways to give the first student his two awards, leaving awards yet to distribute. There are then \binom{3}{2}2111\binom{3}{2}\cdot\binom{5}{2}\cdot\binom{3}{2}\cdot 1=9060+90=\boxed{\textbf{(B) }150}$.
==Solution 2 (variation of Solution 1)== If each student must receive at least one award, then, as in Solution 2, we deduce that the only possible ways to split up the$ (Error compiling LaTeX. ! Missing $ inserted.)53,1,12,2,133\binom{5}{3} = 103223 \cdot 10 \cdot 2 = 60315422\binom{4}{2} = 6223 \cdot 5 \cdot 6 = 9060 + 90 = \textbf{(B) }150$.
==Solution 3== Without the restriction that each student receives at least one award, we could simply take each of the$ (Error compiling LaTeX. ! Missing $ inserted.)533^5=24332^5 = 323 \cdot 32 = 962962153243-96+3=\boxed{\textbf{(B) }150}$.
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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