2006 AIME II 第 4 题

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4.

(a1,a2,a3,,a12)(a_1, a_2, a_3, \ldots, a_{12})(1,2,3,,12)(1, 2, 3, \ldots, 12) 的一个排列,满足 a1>a2>a3>a4>a5>a6a_1 \gt a_2 \gt a_3 \gt a_4 \gt a_5 \gt a_6a6<a7<a8<a9<a10<a11<a12 \begin{aligned} &a_6 \lt a_7 \lt a_8 \lt a_9 \\ &\lt a_{10} \lt a_{11} \lt a_{12} \end{aligned}\text{。}这样的排列的一个例子是 (6,5,4,3,2,1,7,8,9,10,11,12)(6, 5, 4, 3, 2, 1, 7, 8, 9, 10, 11, 12)。求这类排列的个数。

Let (a1,a2,a3,,a12)(a_1, a_2, a_3, \ldots, a_{12}) be a permutation of (1,2,3,,12)(1, 2, 3, \ldots, 12) for which a1>a2>a3>a4>a5>a6a_1 \gt a_2 \gt a_3 \gt a_4 \gt a_5 \gt a_6 and a6<a7<a8<a9<a10<a11<a12. \begin{aligned} &a_6 \lt a_7 \lt a_8 \lt a_9 \\ &\lt a_{10} \lt a_{11} \lt a_{12}. \end{aligned} An example of such a permutation is (6,5,4,3,2,1,7,8,9,10,11,12).(6, 5, 4, 3, 2, 1, 7, 8, 9, 10, 11, 12). Find the number of such permutations.

答案:462
知识点:有限制的排列组合
难度评级:2180
小提示:

a6a_6 小于其他十一个项,所以 a6=1a_6 = 1

a6a_6 is less than all eleven other terms, so a6=1.a_6 = 1.

大提示:

一旦选出剩余 1111 个数中的哪五个填入前五个位置,两半的顺序就都被确定了。

Once you choose which five of the remaining 1111 numbers fill the first five slots, the order of both halves is forced.

解答:

a6a_6 小于这个排列中的其他所有项,所以 a6=1a_6 = 1。现在从剩余 1111 个数中选择五个占据位置 1155:它们必须按递减顺序出现,所以排列方式被确定;其余六个数必须按递增顺序填入位置 771212,这也被确定。

每一种五个数的选择都给出唯一一个合法排列,所以总数为 (115)=462\binom{11}{5} = 462

The term a6a_6 is smaller than every other term of the permutation, so a6=1.a_6 = 1. Now choose which five of the remaining 1111 numbers occupy positions 11 through 5:5: they must appear in decreasing order, so their arrangement is forced, and the other six numbers must fill positions 77 through 1212 in increasing order, which is also forced.

Every choice of the five numbers gives exactly one valid permutation, so the count is (115)=462.\binom{11}{5} = 462.

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