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考题17 2015 AMC 12A

2018-08-22 重点归纳

AMC12是针对高中学生的数学测验，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：

Problem 17

Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?

$AMC12的官方真题$

Solution

Solution 1

We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by $2^8 = 256$ at the end. We casework on how many people are standing.

Case $1:$ $0$ people are standing. This yields $1$ arrangement.

Case $2:$ $1$ person is standing. This yields $8$ arrangements.

Case $3:$ $2$ people are standing. This yields $\dbinom{8}{2} - 8 = 20$ arrangements, because the two people cannot be next to each other.

Case $4:$ $4$ people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding $2$ possible arrangements.

More difficult is:

Case $5:$ $3$ people are standing. First, choose the location of the first person standing ( $8$ choices). Next, choose $2$ of the remaining people in the remaining $5$ legal seats to stand, amounting to $6$ arrangements considering that these two people cannot stand next to each other. However, we have to divide by $3,$ because there are $3$ ways to choose the first person given any three. This yields $\dfrac{8 \cdot 6}{3} = 16$ arrangements for Case $5.$

Alternate Case $5:$ Use complementary counting. Total number of ways to choose 3 people from 8 which is $\dbinom{8}{3}$ . Sub-case $1:$ three people are next to each other which is $\dbinom{8}{1}$ . Sub-case $2:$ two people are next to each other and the third person is not $\dbinom{8}{1}$ $\dbinom{4}{1}$ . This yields $\dbinom{8}{3} - \dbinom{8}{1} - \dbinom{8}{1} \dbinom{4}{1} = 16$

Summing gives $1 + 8 + 20 + 2 + 16 = 47,$ and so our probability is $\boxed{\textbf{(A) } \dfrac{47}{256}}$ .

Solution 2

We will count how many valid standing arrangements there are counting rotations as distinct and divide by $256$ at the end. Line up all $8$ people linearly. In order for no two people standing to be adjacent, we will place a sitting person to the right of each standing person. In effect, each standing person requires $2$ spaces and the standing people are separated by sitting people. We just need to determine the number of combinations of pairs and singles and the problem becomes very similar to pirates and gold aka stars and bars aka sticks and stones aka balls and urns.

If there are $4$ standing, there are ${4 \choose 4}=1$ ways to place them. For $3,$ there are ${3+2 \choose 3}=10$ ways. etc. Summing, we get ${4 \choose 4}+{5 \choose 3}+{6 \choose 2}+{7 \choose 1}+{8 \choose 0}=1+10+15+7+1=34$ ways.

Now we consider that the far right person can be standing as well, so we have ${3 \choose 3}+{4 \choose 2}+{5 \choose 1}+{6 \choose 0}=1+6+5+1=13$ ways

Together we have $34+13=47$ , and so our probability is $\boxed{\textbf{(A) } \dfrac{47}{256}}$ .

Solution 3

We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by $2^8 = 256$ at the end. If we suppose for the moment that the people are in a line, and decide from left to right whether they sit or stand. If the leftmost person sits, we have the same number of arrangements as if there were only $7$ people. If they stand, we count the arrangements with $6$ instead because the person second from the left must sit. We notice that this is the Fibonacci sequence, where with $1$ person there are two ways and with $2$ people there are three ways. Carrying out the Fibonacci recursion until we get to $8$ people, we find there are $55$ standing arrangements. Some of these were illegal however, since both the first and last people stood. In these cases, both the leftmost and rightmost two people are fixed, leaving us to subtract the number of ways for $4$ people to stand in a line, which is $8$ from our sequence. Therefore our probability is $\frac{55 - 8}{256} = \boxed{\textbf{(A) } \dfrac{47}{256}}$

Solution 4

We will count the number of valid arrangements and then divide by $2^8$ at the end. We proceed with casework on how many people are standing.

Case $1:$ $0$ people are standing. This yields $1$ arrangement.

Case $2:$ $1$ person is standing. This yields $8$ arrangements.

Case $3:$ $2$ people are standing. To do this, we imagine having 6 people with tails in a line first. Notate "tails" with $T$ . Thus, we have $TTTTTT$ . Now, we look to distribute the 2 $H$ 's into the 7 gaps made by the $T$ 's. We can do this in $amc数学竞赛官网$ ways. However, note one way does not work, because we have two H's at the end, and the problem states we have a table, not a line. So, we have $AMC12数学竞赛试题及答案$ arrangements.

Case $4:$ $3$ people are standing. Similarly, we imagine 5 $T$ 's. Thus, we have $TTTTT$ . We distribute 3 $H$ 's into the gaps, which can be done ${6 \choose 3}$ ways. However, 4 arrangements will not work. (See this by putting the H's at the ends, and then choosing one of the remaining 4 gaps: ${4 \choose 1}$ =4) Thus, we have ${6 \choose 3}-4=16$ arrangements.