1.In how many ways can three different numbers be selected from the thirty 1,2,3........,30 such that their sum is divisible by 2
2.Caclculate the numbers of ways in which six people can form
(a) a queue (i.e a single file) of six people
(b)a queue of two people and another queue of four people
(c)a group of two people and another group of four people .
(Assume that the order within a group is not significant)
Question about combinations
2010-12-20 3:10 am
回答 (2)
2010-12-24 4:18 am
✔ 最佳答案
(1) If the sum is divisible by 2 then the 3 numbers are even-even-even or even-odd-oddProbability = [15C3 + (15C1)(15C2)] / 30C3 = 0.5(2) (a) There are 6! = 720 ways(b) This is the same as (a). Simply cut the queue into 2, the first 2 people to one queue and the remaining 4 to another queue. Hence the number of ways is still 720(c) Choose 2 people from among 6 to form a group. There are 6C2 = 15 waysThe remaining 4 people automatically fall into the other group
2010-12-20 3:39 am
1 Divide the number in 3 groups:A : 3n [3, 6, 9 ... 30] count = 10B : 3n - 1 [2, 5, 8, ... 29] count = 10C : 3n + 1 [1, 4, 7, ... 28] count = 10For the sum to be divisible by 3, the 3 numbers must be AAA, BBB, CCC or ABC
The number of ways is C(10,3) + C(10,3) + C(10,3) + C(10,1)*C(10,1)*C(10,1)
= 120 + 120 + 120 + 1000
= 1360
2(a) 6!=720
(b) C(6,2)P(2,2)P(4,4)=15*2*24=720
(c) C(6,2)=30
The number of ways is C(10,3) + C(10,3) + C(10,3) + C(10,1)*C(10,1)*C(10,1)
= 120 + 120 + 120 + 1000
= 1360
2(a) 6!=720
(b) C(6,2)P(2,2)P(4,4)=15*2*24=720
(c) C(6,2)=30
收錄日期: 2021-04-26 19:16:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101219000051KK00976