AMC12 2006 B

AMC12 2006 B · Q9

AMC12 2006 B · Q9. It mainly tests Combinations, Parity (odd/even).

How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?

有多少个偶数的三位整数满足其数字从左到右严格递增？

(A) 21 21

(B) 34 34

(D) 72 72

(E) 150 150

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let the integer have digits $a$, $b$, and $c$, read left to right. Because $1 \leq a<b<c$, none of the digits can be zero and $c$ cannot be 2. If $c=4$, then $a$ and $b$ must each be chosen from the digits 1, 2, and 3. Therefore there are $\binom{3}{2}=3$ choices for $a$ and $b$, and for each choice there is one acceptable order. Similarly, for $c=6$ and $c=8$ there are, respectively, $\binom{5}{2}=10$ and $\binom{7}{2}=21$ choices for $a$ and $b$. Thus there are altogether $3+10+21=\boxed{34}$ such integers. (Edited by HMSSONI82)

设该整数从左到右的数字为 $a$、$b$、$c$。由于 $1 \leq a<b<c$，各位数字都不能为 0，且 $c$ 不能为 2。若 $c=4$，则 $a$ 和 $b$ 必须从数字 1、2、3 中选取。因此 $a$ 和 $b$ 的选择有 $\binom{3}{2}=3$ 种，并且每种选择只有一种符合要求的顺序。类似地，当 $c=6$ 和 $c=8$ 时，$a$ 和 $b$ 的选择分别有 $\binom{5}{2}=10$ 和 $\binom{7}{2}=21$ 种。因此总共有 $3+10+21=\boxed{34}$ 个这样的整数。 (Edited by HMSSONI82)

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