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AIME 1990 · 第 8 题

AIME 1990 — Problem 8

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules:

  1. The marksman first chooses a column from which a target is to be broken.

  2. The marksman must then break the lowest remaining target in the chosen column.

If the rules are followed, in how many different orders can the eight targets be broken?

AIME diagram

解析

Solution

Clearly, the marksman must shoot the left column three times, the middle column two times, and the right column three times.

From left to right, suppose that the columns are labeled L,M,L,M, and R,R, respectively. We consider the string LLLMMRRR:LLLMMRRR:

Since the letter arrangements of LLLMMRRRLLLMMRRR and the shooting orders have one-to-one correspondence, we count the letter arrangements:

(83,2,3)=8!3!⋅2!⋅3!=560.\binom{8}{3,2,3} = \frac{8!}{3!\cdot2!\cdot3!} = \boxed{560}. ~Azjps (Solution)

~MRENTHUSIASM (Revision)

Remark

We can count the letter arrangements of LLLMMRRRLLLMMRRR in two ways:

  1. There are 8!8! ways to arrange 88 distinguishable letters. However:

    • Since there are 33 indistinguishable LL's, we have counted each distinct arrangement of the LL's for 3!3! times.

    • Since there are 22 indistinguishable MM's, we have counted each distinct arrangement of the MM's for 2!2! times.

    • Since there are 33 indistinguishable RR's, we have counted each distinct arrangement of the RR's for 3!3! times.

    By the Multiplication Principle, we have counted each distinct arrangement of LLLMMRRRLLLMMRRR for 3!⋅2!⋅3!3!\cdot2!\cdot3! times.

    As shown in the Solution section, we use division to fix the overcount. The answer is 8!3!⋅2!⋅3!=560.\frac{8!}{3!\cdot2!\cdot3!} = 560.

    Alternatively, we can use a multinomial coefficient to obtain the answer directly: (83,2,3)=8!3!⋅2!⋅3!=560.\binom{8}{3,2,3}=\frac{8!}{3!\cdot2!\cdot3!} = 560.

  2. First, we have (83)\binom83 ways to choose any 33 from the 88 positions for the LL's.

    Next, we have (52)\binom52 ways to choose any 22 from the 55 remaining positions for the MM's.

    Finally, we have (33)\binom33 way to choose 33 from the 33 remaining positions for the RR's.

    By the Multiplication Principle, the answer is (83)(52)(33)=560.\binom83\binom52\binom33=560.

~MRENTHUSIASM

Video Solution by OmegaLearn

https://youtu.be/RldWnL4-BfI?t=448

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=NGfMLCRUs3c&t=7s ~ MathEx

Video Solution

https://www.youtube.com/watch?v=eq50QuGasg0