返回题库

HMMT 二月 2006 · TEAM1 赛 · 第 1 题

HMMT February 2006 — TEAM1 Round — Problem 1

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [15] For this problem, our lawn consists of a row of n clumps of grass. This row runs in an east-westdirection. In our formation, each mobot may be oriented toward the north, south, east, or west. Oneexample of an allowable formation if n = 6 is symbolized below:
    ↑ ↑
    · · ←
    ↓
    (The mobot on the third clump will move westward, mowing the first three clumps. Each of the lastthree clumps is mowed by a different mobot.) Here’s another allowable formation for n = 6, considereddifferent from the first:
    ↑ ↑
    · · ← →
    Compute the number of different allowable formations for any given n .
解析

英文解析

  1. [15] For this problem, our lawn consists of a row of n clumps of grass. This rowruns in an east-west direction. In our formation, each mobot may be oriented towardthe north, south, east, or west. One example of an allowable formation if n = 6 issymbolized below:
    ↑ ↑
    · · ←
    ↓
    (The mobot on the third clump will move westward, mowing the first three clumps.
    Each of the last three clumps is mowed by a different mobot.) Here’s another allowableformation for n = 6, considered different from the first:
    ↑ ↑
    · · ← →
    Compute the number of different allowable formations for any given n .
    Solution: Let a be the number of clumps mowed by a mobot oriented to go west,
    and let b be the number of clumps mowed by a mobot oriented to go east. (So in thetwo examples given, ( a, b ) would be (3 , 0) and (3 , 1), respectively.) We may ask howmany allowable formations with a given ordered pair ( a, b ) there are, and then sumour answers up over all possible ordered pairs ( a, b ).
    Given any particular ( a, b ), first of all, a and b had better be non-negative integerssumming to at most n . As long as that’s true, there will be n − a − b clumps of grasswhich are each mowed by a single mobot oriented to go either north or south: theren − a − b 1
    are thus 2 possibilities. So our answer isn n − a n n
    ∑ ∑ ∑ ∑
    n − a − b n − a n − a − 1 0 n − a +1
    2 = (2 + 2 + · · · + 2 ) = (2 − 1)
    a =0 b =0 a =0 a =0
    ∑nn − a +1 n +1 n n − 1 1 = − ( n + 1) + 2 = − ( n + 1) + (2 + 2 + 2 + · · · + 2 )
    a =0
    n +2 n +2 = − ( n + 1) + (2 − 2) = 2 − n − 3 .