Jai 和他的妻子 Kaz 正在举办聚会。Jai 邀请了五位朋友，每位朋友都将带上自己的妻子。

(a) 聚会开始时，这 12 人将站成一排拍照。

(i) 若 Jai 与 Kaz 相邻，且每位朋友都与自己的妻子相邻，这 12 人有多少种不同的排列？

(ii) 若 Jai 和 Kaz 占据队列正中间的两个位置，Jai 的五位朋友站在他们一侧，而这五位朋友的妻子站在另一侧，这 12 人有多少种不同的排列？

(b) 为进行聚会中的一项比赛，将这 12 人随机分为三组，人数分别为 5 人、4 人和 3 人。

求 Jai 和 Kaz 被分在同一组的概率。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

六对夫妻各为一个块，六块排列再乘每对内部两种次序。

Treat each married couple as one block. There are $6!$ orders for the six blocks, and each has 2 internal orders. Thus

另一种生成排列的方式是从左向右选下一位，他或她的配偶随即固定在旁边，剩余可选人数依次减2。

Choose the first person in 12 ways; their spouse must occupy the second position. Choose the third person from the remaining 10; their spouse must occupy the fourth. Continue in this way. Each valid line is generated exactly once, giving

中间两人有2种次序；男性朋友站左或右另有2种，五名朋友与五位妻子分别排列。

Jai and Kaz occupy positions 6 and 7 in 2 orders. The five friends may be on either the left or the right, giving another factor of 2. Arrange the five friends and five wives in their respective sets of positions in $5!$ ways each. Thus

将12个位置分成5、4、3三组，按Jai与Kaz同在每一组分别算概率后相加。

Assign the twelve people uniformly to groups of sizes 5, 4 and 3. The probabilities that Jai and Kaz both occupy each group are respectively

These events are disjoint, so

也可以计完整三组分法，按夫妻同在5、4、3人组分类计数。

The total number of divisions is

If Jai and Kaz are both in the group of 5, choose their other three group members and then the group of 4: $\binom{10}{3}\binom74=4200$ divisions.

If they are both in the group of 4, choose their other two members and then the group of 5: $\binom{10}{2}\binom85=2520$ divisions.

If they are both in the group of 3, choose the remaining member and then the group of 5: $\binom{10}{1}\binom95=1260$ divisions.

Hence