一个飞镖俱乐部有 12 名成员，由 7 名男性和 5 名女性组成。

每星期一，随机选出一支 4 人队伍，代表俱乐部参加比赛。

(a) 求在某个星期一选出的队伍恰好由 1 名男性和 3 名女性组成的概率。

每星期二，飞镖俱乐部选出三支 4 人队伍。各队分别在不同的城镇参加比赛。

(b) 在没有任何限制的情况下，共有多少种不同的选队方法？

(c) 若每队必须有至少 1 名男性和至少 1 名女性，共有多少种不同的选队方法？

这 7 名男性排成一列拍照，其中 George 和 Harry 是兄弟。

(d) 若 George 和 Harry 之间恰好有 2 名男性，这 7 名男性共有多少种不同的排列方法？

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。

从7男中选1、5女中选3，除以所有4人队伍选法。

也可以按不放回的选取次序计算。指定男、女、女、女次序后，男性可占四个位置，乘4。

For one specified order MWWW, the probability is $\frac7{12}\frac5{11}\frac4{10}\frac39$.

There are four possible positions for the man, each giving the same probability. Therefore,

不同城镇区分三个队伍，依次选择三队，因此不除以3的阶乘。

Label the three teams by their towns. Choose the first team, then the second from the eight remaining members; the final four form the third team. Hence

每队至少一男一女，三队男性人数只能是3、3、1或3、2、2。先选特殊构成所在的城镇，再依次选两队，第三队由剩余人员确定。

The possible distributions of the seven men among the three town-labelled teams are $(3,3,1)$ and $(3,2,2)$.

For $(3,3,1)$, choose the town receiving one man in 3 ways. From the other two towns in a fixed order, select teams with three men and one woman:

For $(3,2,2)$, choose the town receiving three men in 3 ways, then choose its team and one of the other teams in a fixed town order:

The remaining members form the final team in each case. Thus

也可以从全部分队中减去含全男队的情况。若有全女队，剩下只有一位女生，另两队中必有全男队，因此只需排除全男队；全男队也不可能有两个，因为只有7名男性。

A team fails the restriction if it is all male or all female.

If a team is all female, only one woman remains for the other two teams, so one of those must be all male. Also, there cannot be two all-male teams because there are only seven men. Thus each invalid division has exactly one all-male team.

Choose its town in 3 ways, its members in $\binom74$ ways, then choose the team for the first of the other two towns from the remaining eight people in $\binom84$ ways. Therefore,

George与Harry的位置相差3，较左一位有四个位置，两兄弟可交换次序。其余五人自由排列。

The two brothers’ position pairs are $(1,4),(2,5),(3,6),(4,7)$.

There are 4 choices of pair, 2 orders for George and Harry, and $5!$ arrangements of the other men. Thus

另一条路线将两兄弟和中间两人合为一块，先选排中间两人，再排兄弟方向及块与剩余三人。

Choose and order the two men between the brothers in $\,{}_5\mathrm{P}_2$ ways. There are 2 possible orders of the brothers.

Treat this four-man string as one block. The block and the three remaining men have $4!$ arrangements. Hence