AMC10 2005 B

AMC10 2005 B · Q21

AMC10 2005 B · Q21. It mainly tests Basic counting (rules of product/sum), Combinations.

Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b \ne a$. What is the value of $q/p$?

帽子里放入40张纸条，每张纸条上写着数字1、2、3、4、5、6、7、8、9或10，每个数字有四张纸条。从帽子里随机无放回抽取四张纸条。设$p$为四张纸条上数字都相同的概率。设$q$为其中两张纸条上数字为$a$，另外两张为$b\ne a$的概率。$q/p$的值是多少？

(A) 162 162

(B) 180 180

(D) 360 360

(E) 720 720

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A) The total number of ways that the numbers can be chosen is $\binom{40}{4}$. Exactly 10 of these possibilities result in the four slips having the same number. Now we need to determine the number of ways that two slips can have a number $a$ and the other two slips have a number $b$, with $b \ne a$. There are $\binom{10}{2}$ ways to choose the distinct numbers $a$ and $b$. For each value of $a$ there are $\binom{4}{2}$ to choose the two slips with $a$ and for each value of $b$ there are $\binom{4}{2}$ to choose the two slips with $b$. Hence the number of ways that two slips have some number $a$ and the other two slips have some distinct number $b$ is $$ \binom{10}{2}\cdot\binom{4}{2}\cdot\binom{4}{2}=45\cdot6\cdot6=1620. $$ So the probabilities $q$ and $p$ are $\frac{10}{\binom{40}{4}}$ and $\frac{1620}{\binom{40}{4}}$, respectively, which implies that $$ \frac{p}{q}=\frac{1620}{10}=162. $$

（A）选择这些数字的总方法数是 $\binom{40}{4}$。其中恰有 10 种情况会使四张纸条上的数字相同。现在需要确定这样的情况数：两张纸条上的数字为 $a$，另外两张纸条上的数字为 $b$，且 $b \ne a$。选择不同的数字 $a$ 和 $b$ 有 $\binom{10}{2}$ 种方法。对每个 $a$，选出两张写 $a$ 的纸条有 $\binom{4}{2}$ 种方法；对每个 $b$，选出两张写 $b$ 的纸条也有 $\binom{4}{2}$ 种方法。因此，两张纸条为某个数字 $a$ 且另外两张为不同数字 $b$ 的方法数为 $$ \binom{10}{2}\cdot\binom{4}{2}\cdot\binom{4}{2}=45\cdot6\cdot6=1620. $$ 所以概率 $q$ 和 $p$ 分别为 $\frac{10}{\binom{40}{4}}$ 和 $\frac{1620}{\binom{40}{4}}$，从而 $$ \frac{p}{q}=\frac{1620}{10}=162. $$

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