Question

在 Aimeville 的 $900$ 位居民中，有 $195$ 人拥有钻戒， $367$ 人拥有一套高尔夫球杆， $562$ 人拥有一把园艺铲。此外，这 $900$ 位居民每人都拥有一袋心形糖果。有 $437$ 位居民恰好拥有这些物品中的两样，有 $234$ 位居民恰好拥有这些物品中的三样。求 Aimeville 中拥有全部四样物品的居民人数。

Among the $900$ residents of Aimeville, there are $195$ who own a diamond ring, $367$ who own a set of golf clubs, and $562$ who own a garden spade. In addition, each of the $900$ residents owns a bag of candy hearts. There are $437$ residents who own exactly two of these things, and $234$ residents who own exactly three of these things. Find the number of residents of Aimeville who own all four of these things.

答案

答案：73

知识点：容斥原理韦恩图双重计数

难度评级：2010

解答：

将四种物品的拥有人数相加，得到 $195 + 367 + 562 + 900 = 2024$ 次物品拥有记录，来自 $900$ 位居民。因为每人都有一袋心形糖果，每位居民至少拥有一件物品；若某位居民恰好拥有 $k$ 件物品，那么除第一次计数外还会多计 $k - 1$ 次。

设有 $n_4$ 位居民拥有全部四样物品，则多出的计数总数为所以 $1124 = 905 + 3 n_4$ ，得到 $n_4 = \frac{219}{3} = 73$ 。 $\begin{aligned} 2024 - 900 &= 437 \cdot 1 + 234 \cdot 2 \\ &\quad {}+ n_4 \cdot 3, \end{aligned}$

Adding the four ownership counts gives $195 + 367 + 562 + 900 = 2024$ item ownerships among the $900$ residents. Since everyone owns a bag of candy hearts, every resident owns at least one item, and a resident owning exactly $k$ items is counted $k - 1$ times beyond the first.

If $n_4$ residents own all four things, the extra counts total $\begin{aligned} 2024 - 900 &= 437 \cdot 1 + 234 \cdot 2 \\ &\quad {}+ n_4 \cdot 3, \end{aligned}$ so $1124 = 905 + 3 n_4,$ giving $n_4 = \frac{219}{3} = 73.$

2024 AIME II 第 1 题

答案：73

难度评级：2010

解答：

其他年份的第 1 题