可以有解釋給我嗎???唔該哂~~~我不知道幾時要用Permutations和Combinations,用乘或者加??~~
1a)In how many ways can 9 students be divided into 3 groups of 2,3 and 4 respectively?
b)In how many ways can 9 students be divided into groups of 3 if no regard is paid to the order of the group?
2.A committee of 3 is to be chosen from 4 married couples.Find how many ways can the committee be chosen if
a)the committee must consist of one woman and two men,
b)all are eligible except that any couple cannot both serve on the committee.
3a)In how many ways can 10 basketball players be divided into two teams of 5 players each?
b)In how many ways can 10 basketball players be divided into two teams of 5 players so that the 2 best players are on opposite teams?
Permutations,Combinations???
2009-09-21 1:27 am
回答 (1)
2009-09-21 6:51 am
✔ 最佳答案
(1)(a) We first choose 2 from 9 to make up the first group. There is 9C2 ways. Then choose 3 from the remaining 7 to make the second group. There are 7C3 ways. The remaining 4 will be the last group. So there are in total 9C2 x 7C3 = 36 * 35 = 1260 ways.(b) I guess the meaning of the question is group A = 3 people, B = 3 people and C = 3 people.
There are 9C3 (84) ways to build the first group, and 6C3 (20) ways to build the second. The remaining 3 becomes a group. So there are 84 * 20 = 1680 ways. However, if the order of the group is not important, then there are 1680 / 3! = 280 ways.
(2) (a) We choose 1 woman from among 4 (4C1) and 2 men from among 4 (4C2) so there are 4 x 6 = 24 ways.
(b) There are 8C3 (56) ways to choose 3 members from 8 people.
For the case when a couple is in the committee, there are 4C1 ways to select one couple out of 4, and 6C1 ways to select the remaining member from the remaining 6 people. So the total number of ways to have the committee having a couple is 4 * 6 = 24.
The number of ways with no couple in the committee = 56 - 24 = 32.
(3) (a) We choose 5 people from 10 to build the first team X. There are 10C5 ways = 252 ways. There is no need choose for the second team Y. Disregard the order of the 2 teams, (ABCDE in X and FGHIJ in Y is the same as ABCDE in Y and FGHIJ in X) there are 252 / 2 = 126 ways.
(b) Let the best player A be in team X, and 2nd best player B in team Y. We choose 4 out of 8 to be in group X. There are 8C4 ways = 70 ways.
收錄日期: 2021-04-23 23:18:22
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