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排列组合头痛

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楼主
发表于 2013-7-15 11:53:52 | 只看该作者 回帖奖励 |倒序浏览 |阅读模式
经常看见排列组合答案的解释,分母是3!, 想问一下这表示什么意思?我从来就没弄明白分母这个表达的意思
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沙发
发表于 2013-7-15 12:02:46 | 只看该作者
阶乘啊兄弟……
板凳
发表于 2013-7-15 12:05:47 | 只看该作者
3*2*1= 3!     n!=n*(n-1)*(n-2)*....*2*1   ~ 兄弟加油复习。。。
地板
 楼主| 发表于 2013-7-15 22:08:14 | 只看该作者
我知道,我不是这个意思,我把那道题目贴出来
5#
 楼主| 发表于 2013-7-15 23:50:19 | 只看该作者
站着 发表于 2013-7-15 12:02
阶乘啊兄弟……

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

This Combinations problem is asking for the number of ways to select 3 cars from 8 (each of the 2 models comes in 4 different colors for a total of 2 x 4 = 8 different types of cars) with the restriction that none of the selected cars be the same color.
We can treat the Carson family’s purchase as a sequence of decisions: the Carsons can initially purchase any one of the 8 cars, but once they have chosen the first car, their choice for the subsequent purchases is limited. This type of choice decision fits well into the Slot Method.
For the first choice, the family can choose from all 8 cars. After they have selected the first vehicle, they have fewer choices for the second pick because they cannot select another car of the same color. For example, if the family purchases a green Model A they cannot also purchase a green Model B. Therefore, we have eliminated both a green Model A and a green Model B from the second choice, leaving only 6 cars from which to choose.

A similar scenario occurs after the second choice, leaving only 4 cars for the final choice. Multiplying these choices together to get the total number of choices we have 8×6×4. (Don’t multiply this out yet! Save yourself some trouble by simplifying first.)

The order in which the purchases are made is not important so we must divide by the factorial of the number of choices to eliminate over-counting:

8*6*4/ 3!=32
6#
 楼主| 发表于 2013-7-15 23:57:36 | 只看该作者
xiqiong 发表于 2013-7-15 12:05
3*2*1= 3!     n!=n*(n-1)*(n-2)*....*2*1   ~ 兄弟加油复习。。。

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

This Combinations problem is asking for the number of ways to select 3 cars from 8 (each of the 2 models comes in 4 different colors for a total of 2 x 4 = 8 different types of cars) with the restriction that none of the selected cars be the same color.
We can treat the Carson family’s purchase as a sequence of decisions: the Carsons can initially purchase any one of the 8 cars, but once they have chosen the first car, their choice for the subsequent purchases is limited. This type of choice decision fits well into the Slot Method.
For the first choice, the family can choose from all 8 cars. After they have selected the first vehicle, they have fewer choices for the second pick because they cannot select another car of the same color. For example, if the family purchases a green Model A they cannot also purchase a green Model B. Therefore, we have eliminated both a green Model A and a green Model B from the second choice, leaving only 6 cars from which to choose.

A similar scenario occurs after the second choice, leaving only 4 cars for the final choice. Multiplying these choices together to get the total number of choices we have 8×6×4. (Don’t multiply this out yet! Save yourself some trouble by simplifying first.)

The order in which the purchases are made is not important so we must divide by the factorial of the number of choices to eliminate over-counting:

8*6*4/ 3!=32

我不明白为啥是3!
7#
发表于 2013-7-16 11:02:01 | 只看该作者
3!就是P(3,3)
8x6x4其实是排列,不是组合,也就是说,你8x6x4个可能里包含了购买次序不同但结果相同的三辆车的可能。
因为每种组合都会有p(3,3)次重复的可能性,所以要除掉。

只要不能直接用C和P计算的题,都得考虑计算结果是排列还是组合,有没有重复。
希望对你有帮助〜
8#
发表于 2013-7-16 11:06:53 | 只看该作者
对了,多插一句,C和P的区别从公式上就能体会到,就差一个拿球次数的阶乘。
9#
 楼主| 发表于 2013-7-16 11:42:09 | 只看该作者
站着 发表于 2013-7-16 11:06
对了,多插一句,C和P的区别从公式上就能体会到,就差一个拿球次数的阶乘。 ...

我愚笨,兄台。是不是除以P3就是去除重复的一个公式还是什么?
我无法理解为什么是除以p
是公式吗?
10#
发表于 2013-7-16 13:47:40 | 只看该作者
1.2.3
1.3.2
2.1.3
2.3.1
3.1.2
3.2.1
这6种情况都是1和2和3,重复的。
重复的次数就是p(3,3)次,所以除掉。
只能解释到这了……
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