AMC12 2022 A

AMC12 2022 A · Q10

AMC12 2022 A · Q10. It mainly tests Permutations, Casework.

How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?

有几种方法将整数$1$到$14$分成$7$对，使得每对中较大的数至少是较小数的$2$倍？

(A) 108 108

(B) 120 120

(D) 132 132

(E) 144 144

Answer

Correct choice: (E)

正确答案：（E）

Solution

Clearly, the integers from $8$ through $14$ must be in different pairs, so are the integers from $1$ through $7.$ Note that $7$ must pair with $14.$ We pair the numbers $1,2,3,4,5,6$ with the numbers $8,9,10,11,12,13$ systematically: - $6$ can pair with either $12$ or $13.$ - $5$ can pair with any of the three remaining numbers from $10,11,12,13.$ - $1,2,3,4$ can pair with the other four remaining numbers from $8,9,10,11,12,13$ without restrictions. Together, the answer is $2\cdot3\cdot4!=\boxed{\textbf{(E) } 144}.$

显然，$8$到$14$的整数必须成不同对，$1$到$7$也是如此。注意$7$必须与$14$配对。我们系统地将$1,2,3,4,5,6$与$8,9,10,11,12,13$配对： - $6$可与$12$或$13$配对，即$2$种。 - $5$可与剩余的$10,11,12,13$中任三个，即$3$种。 - $1,2,3,4$可与剩余的四个数任意配对，即$4!$种。总计 $2\cdot3\cdot4!=\boxed{\textbf{(E) } 144}$。

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