標題:
S5 math Probability ex plz
發問:
- 到達墾丁的票價
- 初音未來街機IC卡
- 台北101的簡介
- 急到爆炸 劍湖山與月眉10年來的人數統計
- 二代健保對於失業的單身族如何計算-
- 2006觀護盃
- 嘉義長庚科技大學 這樣有機會嗎-
- 中部哪裡有地方可以滑草-
- 台式推拿~~和泰式哪個比較好
- 台北車站到民權公園
此文章來自奇摩知識+如有不便請留言告知
Q1. 6 guests A,B,C,D,E and F are arranged to sit in two row. Each row has 3 seats. How many possible arrangements are there ifa) A and B must sit in different rows? (ans: 432)b)A and B must sit in the same row and C must sit in another row? (ans: 216)Q2.10 people are going to travel with two cars. One car... 顯示更多 Q1. 6 guests A,B,C,D,E and F are arranged to sit in two row. Each row has 3 seats. How many possible arrangements are there if a) A and B must sit in different rows? (ans: 432) b)A and B must sit in the same row and C must sit in another row? (ans: 216) Q2.10 people are going to travel with two cars. One car can carry 7 people and the other car can carry 5 people. In how many ways can the 10 people be arranged into two cars and the 2 drivers be assigned if a) all of them have driving licenses (ans: 13 860) b) only 3 of them have driving licenses (ans: 924) Q3. A staff number of a company consists of 2 letters from A to E, followed by 4-digits from 0 to 9. If no repeated letters and digit are allowed, find the number of possible staff numbers (ans: 100 800) Q4 In a box, there are 10 balls numbered 1 to 10. 4 balls are selected from the box. In how may ways can the selected if the greatest number is 7? (ans: 20) Q5. There are 10 distinct points on a coordinate plane. If no three points are collinear , find the number of lines that can be drawn by joining these points (ans: 45)
最佳解答:
1. (a) Arrange A to take a seat in one of the 2 rows. (2C1 x 3P1) Arrange B to take a seat in another row. (3P1) Arrange the rest four guests to take the rest 4 seats. (4P4) No. of possible arrangements = 2C1 x 3P1 x 3P1x 3P3 = 2 x 6 x 6 x 6 = 432 (b) Arrange A and B to take 2 seats in one the 2 rows. (2C1 x3P2) Arrange C to take a seat in another row. (3-P1) Arrange the rest 3 guests to take the rest 3 seats. (3P3) No. of possible arrangements = 2C1 x 3P2- x 3P1x 3P3 = 2 x 6 x 3 x 6 = 216 ===== 2. (a) Arrange the two drivers. (10P-2) Arrange 6, 5 or 4 people to the car of 7 people. (8C6 + 8C5+ 8C4) The rest people take another car. (1) No. of arrangements = 10P2 x (8C6 + 8C5+ 8C4) = 90 x (28 + 56 + 70) = 13860 (b) Arrange the two drivers. (3P2) All other steps are the same. No. of arrangements = 3P2 x (8C6 + 8C5+ 8C4) = 6 x (28 + 56 + 70) = 924 ===== 3. Arrange two letters from A to E. (5P2) Arrange 4 digits from 0 to 9. (10P4) No. of possible staff numbers = 5P2 x 10P4 = 20 x 5040 = 100800 ===== 4. Take ball numbered 7. (1C1) Take 3 balls numbered from 1 to 6. (6C3) No. of ways = 1C1 x 6C3 = 1 x 20 = 20 ===== 5. Join any 2 points can form a line. No. of lines = 10C2 = 45
其他解答:6FE6251C5F3F8A0F
留言列表
