AMC10 2018 A

AMC10 2018 A · Q11

AMC10 2018 A · Q11. It mainly tests Basic counting (rules of product/sum), Combinations.

When 7 fair standard 6-sided dice are thrown, the probability that the sum of the numbers on the top faces is 10 can be written as $\frac{n}{6^7}$, where $n$ is a positive integer. What is $n$?

掷7个公平的标准的6面骰子，顶面数字之和为10的概率可以写成 $\frac{n}{6^7}$，其中 $n$ 是正整数。$n$ 是多少？

(A) 42 42

(B) 49 49

(D) 63 63

(E) 84 84

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): The only ways to achieve a sum of 10 by adding 7 unordered integers between 1 and 6 inclusive are (i) six 1s and one 4; (ii) five 1s, one 2, and one 3; or (iii) four 1s and three 2s. The number of ways to order the outcomes among the 7 dice are 7 in case (i), $7\cdot 6=42$ in case (ii), and $\binom{7}{3}=35$ in case (iii). There are $6^7$ possible outcomes. Therefore $n=7+42+35=84$.

答案（E）：用 7 个介于 1 到 6（含）之间的无序整数相加得到 10 的方式只有：(i) 六个 1 和一个 4；(ii) 五个 1、一个 2 和一个 3；或 (iii) 四个 1 和三个 2。在 7 个骰子中对这些结果进行排列的方式数分别为：情形 (i) 为 7 种，情形 (ii) 为 $7\cdot 6=42$ 种，情形 (iii) 为 $\binom{7}{3}=35$ 种。共有 $6^7$ 种可能结果。因此 $n=7+42+35=84$。

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