S5 概率 Permutation and Combi...

[匿名]

2013-04-27 6:16 pm

In a graduation ceremony(畢業典禮), 10 seats in two rows of 5 are reserved for the guests. Among the guests, there are 4 women and 6 men. What is the number of seating arrangements if

(a) there are no other restrictions?
(b) a particular guest must be seated on the leftmost in the first row?
(c) all the women are sitting on the same row?
(d) at most one women sit in the first row?

回答 (1)

用戶9112

2013-04-27 11:26 pm

✔ 最佳答案

(a) This is equivalent to arranging 10 people in a row and seat them in sequenceThere are 10!=3628800 ways(b) This is equivalent to arranging the 9 other people similar to (a)Number of ways=9!=362880 (c) If all women are in front row, we need to pick one man to sit with them.There are C(6,1)=6 ways to select this manTotal number of arrangements=6(5!)(5!)=86400Consider also the case when all women are in back row, number of ways=86400×2=172800(d) If there is no woman in front row, there are 86400 ways (see part c)If there is 1 woman in the front row, there are 4 ways to select this womanThen we have to select 4 men out of 6 to sit with her, i.e.C(6,4)=15 waysTotal number of arrangements=(4)(15)(5!)(5!)=864000Hence required answer=86400+864000=950400

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