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标题: Manhattan 模拟数学排列组合问题 [打印本页]

作者: 秦宝子 时间: 2017-6-7 11:38
标题: Manhattan 模拟数学排列组合问题
manhattan 模拟
[size=12.0012px]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?
[size=12.0012px]

[size=12.0012px]答案是32
[size=12.0012px]

[size=12.0012px]我自己做的是16，我想问问各位NN我为什么错，我的思路哪里有问题？：
[size=12.0012px]首先3辆车有4种组合：AAA BBB ABB AAB，然后因为每辆车颜色不能一样，所以在AAA里面有（4*3*2）/3！=4种可能，剩下的BBB,ABB,AAB同理各有4种可能。
[size=12.0012px]最后加起来4+4+4+4=16 看起来没啥问题，但是不知道为什么比答案少了1/2

[size=12.0012px].。。。。
[size=12.0012px]求解答

作者: givemeoffer 时间: 2017-6-7 11:46
答案是32吗？

先选3个不同的颜色，有4种可能性。针对每一种三色方案，又有8种组合方式。所以4x8=32种。（补充：对于上述的8种组合方式，可以假定颜色是1.2.3.4号色；那么在选定1.2.3号色的情况下，就有A1A2A3，A1A2B3，A1B2A3，A1B2B3，B1A2A3，B1A2B3，B1B2A3，B1B2B3总计8种。）

作者: 秦宝子 时间: 2017-6-7 11:51

givemeoffer 发表于 2017-6-7 11:46
答案是32吗？

先选3个不同的颜色，有4种可能性。针对每一种三色方案，又有8种组合方式。所以4x8=32种。（ ...

答案是32
manhattan给的解释：
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！

作者: givemeoffer 时间: 2017-6-7 11:55
是否LZ要重新考虑一下：
4种组合：AAA BBB ABB AAB 其中的每一种组合是否都有且仅有4种可能
或许从model入手不如从color入手那么directly
欢迎补充指正。

作者: 秦宝子 时间: 2017-6-7 11:57

givemeoffer 发表于 2017-6-7 11:46
答案是32吗？

先选3个不同的颜色，有4种可能性。针对每一种三色方案，又有8种组合方式。所以4x8=32种。（ ...

谢谢答复，你的方法也不错，也没问题。但是能帮我分析一下我为什么错啊，以为我先入为主的思路就是这样，以后做题也可能会这样，那这种题型我就会继续错，。。

作者: givemeoffer 时间: 2017-6-7 12:06

秦宝子发表于 2017-6-7 11:57
谢谢答复，你的方法也不错，也没问题。但是能帮我分析一下我为什么错啊，以为我先入为主的思路就是这样， ...

按照你的思路，AAA和BBB分别有4种方案没错。但是AAB和ABB错了。
以AAB为例，应该是1C4*2C3=12种（Combination: 4个号色中选1个作为B的配色，在余下3个号色中选2个作为A的配色）; ABB同理。
因此，就应该是（4+4+12+12）*4=32种。
P.S 我觉得这种思路比较复杂。而直接从题目的设限条件different colors入手比较方便，个人感觉。

作者: 秦宝子 时间: 2017-6-7 12:20

givemeoffer 发表于 2017-6-7 12:06
按照你的思路，AAA和BBB分别有4种方案没错。但是AAB和ABB错了。
以AAB为例，应该是1C4*2C3=12种（Combinat ...

明白了=。=，，看来我排列组合的思想还是不到位，还得深造深造

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