AMC12 2023 B

AMC12 2023 B · Q8

AMC12 2023 B · Q8. It mainly tests Basic counting (rules of product/sum), Combinations.

How many nonempty subsets $B$ of $\{0, 1, 2, 3, \cdots, 12\}$ have the property that the number of elements in $B$ is equal to the least element of $B$? For example, $B = \{4, 6, 8, 11\}$ satisfies the condition.

集合$\{0, 1, 2, 3, \cdots, 12\}$有多少个非空子集$B$，使得$B$的元素个数等于$B$的最小元素？例如，$B = \{4, 6, 8, 11\}$满足条件。

(A) 256 256

(B) 136 136

(D) 144 144

(E) 156 156

Answer

Correct choice: (D)

正确答案：（D）

Solution

There is no way to have a set with 0. If a set is to have its lowest element as 1, it must have only 1 element: 1. If a set is to have its lowest element as 2, it must have 2, and the other element will be chosen from the natural numbers between 3 and 12, inclusive. To calculate this, we do $\binom{10}{1}$. If the set is the have its lowest element as 3, the other 2 elements will be chosen from the natural numbers between 4 and 12, inclusive. To calculate this, we do $\binom{9}{2}$. We can see a pattern emerge, where the top is decreasing by 1 and the bottom is increasing by 1. In other words, we have to add $1 + \binom{10}{1} + \binom{9}{2} + \binom{8}{3} + \binom{7}{4} + \binom{6}{5}$. This is $1+10+36+56+35+6 = \boxed{\textbf{(D) 144}}$.

不可能有0个元素的集合。如果集合的最小元素为1，则它必须只有一个元素：1。如果集合的最小元素为2，则必须包含2，其他元素从3到12（包含）中选择，计算为$\binom{10}{1}$。如果最小元素为3，则其他2个元素从4到12（包含）中选择，计算为$\binom{9}{2}$。可以看到一种模式，上限每次减1，下限每次加1。也就是说，我们需要计算$1 + \binom{10}{1} + \binom{9}{2} + \binom{8}{3} + \binom{7}{4} + \binom{6}{5}$。这是$1+10+36+56+35+6 = \boxed{\textbf{(D) 144}}$。

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