3枝花插在三个不同的花瓶里有多少种组合？ 四枝插三个瓶呢？四枝插四个瓶呢？
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C33 C43 C44好像是吧
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请问：是教多大（或几年级）的孩子？
如果是大人的话就套一下排练组合的公式就可以了。
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请问排练组合的公式？
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楼上的，俺当时总结了这个教的小胖子，大家要是觉得有用就给撒几分哈~~
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万分感谢，这几天正为这事发愁呢。
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高中的题，小学就学了？？
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没办法补习班的题
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这是中学学的东西。我教我家的小学生不用这种方法的。可能我比较小瞧我家的那个小学生。让我回去教教她这种公式试试。
[ 本帖最后由 Holly2010 于 2010-8-17 11:25 编辑 ]
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反正小福州越搞越糊涂了。
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难怪小胖子考入第二名中学，神童啊
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charmaine的附件是高中数学的内容，用来教小学生，我佩服啊 charmaine难怪是个博士家长，敬仰啊。
我的认知，大部分小学生是肯定不能这样教的，套公式其实小朋友还是不懂概念，这些题型是高中数学的问题 - 太超前了。
小学生只有有基本的排列组合概念就可以了，至少是在精英中学的考试中可以应付了。
排列组合里的精髓部分不是公式，而是提炼出来的最基本的两个原理：
- 加法原理
- 乘法原理
小学生对这两个原理有认知和会简单的运用就可以了。
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老朱能讲细点吗小福州求你了。
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问题没有说清楚，
1）这三枝花是否相同
2）每个花瓶是否最多只能插一支花
如果两者都是否定回答的话，算起来就很繁琐了
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啊，真的啊？
OC的孩子年纪太小肯定不行，精英考的孩子应该还好吧
俺就是把那个文件打印回家，让小胖子自己读一遍，再跟他讲一遍，然后出几个题，确定他明白了，就完了，似乎他也没有提出什莫异议
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完全同意. 但是我估计小学的题目应该不会那么难.
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没有最好的只有最适合的，今晚等小福州回来我让他试一下你的方法，如果还不行我 只有撞墙了。
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这是神童天才补习班的题目
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最繁琐的那种情况
我接触的很多鬼佬大学生，估计有一大半都做不出来
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怎么打不开呢? 我回家再试试.是三年纪ICAS里面的题型.
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是.docx 文件，打不开吗？或者我转成.doc?
估计三年级的孩子用我的办法不行，听朱兄的吧。
[ 本帖最后由 charmaine 于 2010-8-17 13:04 编辑 ]
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今天是ICAS数学考试的日子, 昨晚他做了08年三年纪的题, 这是其中一道. 感谢朱斑给我答案.
这道题是三放三的, 我是用图解法一种种画给他的. 但再多点儿就不容易图解了.
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教教我也行呀! 我打不开, 过会儿我回家试试. 先加分!
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见y12345678的疑问，楼主的题目题意不清，没法做。
加法原理简单的说就是分类问题。
比如，做一件事情一共有3种分类，第一类有3种方法，第二类有4种方法，第三类有5种方法，那么做这件事情就有3+4+5=12种方法。
乘法原理可以理解为分步问题。
比如，做一件事情，要分三步，第一步有2个方法，第二步有3个方法，第三步有4个方法，那么做这件事情就是有2x3x4=24种方法。
加法原则的要点是每一类都是相互独立没有关联的，单独做任何一类里的任何一种way都可以完成这件事情。
比如，你去City，可以坐Train每天有2个班次，也可以Bus每天有3个班次，那么去city的way就一共有2+3=5种，因为独立的坐train或者bus的任何一个班次都可以完成这件事（去City），这就是加法原则。
乘法原则的要点是每一步都是想关联的，要分步骤的一步一步才能完成这件事情,是基于加法原则的延伸。
比如，A点到B点有3条路，B点到C点有4条路，那么A点到C点的走法一共有3x4=12种，因为A到B的第一种选择里对应了B到C的4种选择，A到C一定要先B后C才能完成这件事情，重复这样的过程3次（每次都是独立的，每次都完成了这件事情）所以就有4+4+4=3x4=12。
举了例子：
书架上一共有18本书，其中English Book有5本，Maths Book有6本，Science Book有7本？
1. 从书架上任取1本书，有多少种办法？
因为任取一本书就完成了这件事情（无所谓是English，Maths还是Science），那么直接用加法原则，任取English book有5种，Maths有5种，Science有7种，那么一共有5+6+7=18种。
不考精英中学的小学生知道这个就行了。:)
2. 从书架上取English，Maths和Science各一本书，共有多少种取法？
先要明确做这件事情要有三个步骤，第一步 English book里取1本有5种取法，第二步从Maths book里取1本有6种取法，第三步从Science book里取1本有7种取法，那么乘法原理，一共有5x6x7=210种方法。
考精英中学的小学生需要会做这类题。:)
3. 从书架上任取不同类型的书两本，共有多少种取法？
先要明确做这件事情首先要分成3类，第一类是English+Maths各一本，第二类是Maths+Science各一本，第三类是English+Science各一本。
然后在每一类里分步选择 - 乘法原理：
第一类 （English+Maths）
5本English book里取1本有5种，6本maths book里取1本有6种，所以第一类里有5x6=30种。
第二类 （Maths+Science）
6本Maths book里取1本有6种，7本Science book里取1本有7种，所以第二类里有6x7=42种。
第三类 （English+Science）
5本English book里取1本有5种，7本science book里取1本有7种，所以第三类里有5x7=35种。
前面的是分类问题，所以要用加法原则：
第一类+第二类+第三类，30+42+35=107种。
考JR和Baulkhill High和全奖的大概需要熟练掌握这类的。呵呵，因为这是加法原则和乘法原则的混合运用。:)
小学生的题型我个人觉得也就仅限于这两种原理的一般运用了，不会很复杂，高中的数学公式肯定是用不上的。我个人认为这些公式让小学生学危害很大。我知道有很多高中生学了排列组合的公式，你让他做上面这个例题的第三道，他可能还是不会做，因为基本原理不清。
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谢谢茶面，可惜是新手不能加分。真的selective应该没这么复杂吗？象今年的year5数学竞赛有这样一题，4个小孩不知道大小，有24种可能（老大-老四）如果是5个小孩的话会增加几种可能？这个就简单多了，我教过我丫头排立组合的公式，到了她告诉我还是硬排出来的，忘了如何套。不知selective和icas哪个难？
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非常感谢回头一定加分
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老朱我已前用这个方法教混了儿子 Easy Permutations and Combinations
I’ve always confused “permutation” and “combination” — which one’s which?
Here’s an easy way to remember: permutation sounds complicated, doesn’t it? And it is. With permutations, every little detail matters. Alice, Bob and Charlie is different from Charlie, Bob and Alice (insert your friends’ names here).
Combinations on the other hand, are pretty easy going. The details don’t matter. Alice, Bob and Charlie is the same as Charlie, Bob and Alice.
Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter).
Permutations: The hairy details
Let’s start with permutations, or all possible ways of doing something. We’re using the fancy-pants term “permutation”, so we’re going to care about every last detail, including the order of items. Let’s say we have 8 people:
1: Alice
2: Bob
3: Charlie
4: David
5: Eve
6: Frank
7: George
8: Horatio
How many ways can we pick a Gold, Silver, and Bronze medal for “Best friend in the world”?
We’re going to use permutations since the order we hand out these medals matters. Here’s how it breaks down:
• Gold medal: 8 choices: A B C D E F G H (Clever how I made the names match up with letters, eh?). Let’s say A wins the Gold.
• Silver medal: 7 choices: B C D E F G H. Let’s say B wins the silver.
• Bronze medal: 6 choices: C D E F G H. Let’s say… C wins the bronze.
We picked certain people to win, but the details don’t matter: we had 8 choices at first, then 7, then 6. The total number of options was 8 * 7 * 6 = 336.
Let’s look at the details. We had to order 3 people out of 8. To do this, we started with all options (8) then took them away one at a time (7, then 6) until we ran out of medals.
We know the factorial is:
Unfortunately, that does too much! We only want 8 * 7 * 6. How can we “stop” the factorial at 5?
This is where permutations get cool: notice how we want to get rid of 5*4*3*2*1. What’s another name for this? 5 factorial!
So, if we do 8!/5! we get:
And why did we use the number 5? Because it was left over after we picked 3 medals from 8. So, a better way to write this would be:
where 8!/(8-3)! is just a fancy way of saying “Use the first 3 numbers of 8!”. If we have n items total and want to pick k in a certain order, we get:
just means “Use the first k numbers of n!”
And this is the fancy permutation formula: You have n items and want to find the number of ways k items can be ordered:
Combinations, Ho!
Combinations are easy going. Order doesn’t matter. You can mix it up and it looks the same. Let’s say I’m a cheapskate and can’t afford separate Gold, Silver and Bronze medals. In fact, I can only afford empty tin cans.
How many ways can I give 3 tin cans to 8 people?
Well, in this case, the order we pick people doesn’t matter. If I give a can to Alice, Bob and then Charlie, it’s the same as giving to Charlie, Alice and then Bob. Either way, they’re going to be equally disappointed.
This raises an interesting point — we’ve got some redundancies here. Alice Bob Charlie = Charlie Bob Alice. For a moment, let’s just figure out how many ways we can rearrange 3 people.
Well, we have 3 choices for the first person, 2 for the second, and only 1 for the last. So we have 3 * 2 * 1 ways to re-arrange 3 people.
Wait a minute… this is looking a bit like a permutation! You tricked me!
Indeed I did. If you have N people and you want to know how many arrangements there are for all of them, it’s just N factorial or N!
So, if we have 3 tin cans to give away, there are 3! or 6 variations for every choice we pick. If we want to figure out how many combinations we have, we just create all the permutations and divide by all the redundancies. In our case, we get 336 permutations (from above), and we divide by the 6 redundancies for each permutation and get 336/6 = 56.
The general formula is
which means “Find all the ways to pick k people from n, and divide by the k! variants”. Writing this out, we get our combination formula, or the number of ways to combine k items from a set of n:
A few examples
Here’s a few examples of combinations (order doesn’t matter) from permutations (order matters).
• Combination: Picking a team of 3 people from a group of 10. C(10,3) = 10!/(7! * 3!) = 10 * 9 * 8 / (3 * 2 * 1) = 120.
Permutation: Picking a President, VP and Waterboy from a group of 10. P(10,3) = 10!/7! = 10 * 9 * 8 = 720.
• Combination: Choosing 3 desserts from a menu of 10. C(10,3) = 120.
Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P(10,3) = 720.
Don’t memorize the formulas – it’s better to know why they work. Combinations sounds simpler than permutations, and they are. You have fewer combinations than permutations.
Navigate a Grid Using Combinations And Permutations
Puzzles can help develop your intuition -- figuring how to navigate a grid helped me understand combinations and permutations.
Suppose you're on a 4 × 6 grid, and want to go from the bottom left to the top right. How many different paths can you take? Avoid backtracking -- you can only move right or up.
Spend a few seconds thinking about how you'd figure it out.
Insight: Convert Pictures To Text
When considering the possible paths (tracing them out with your finger), you might whisper "Up, right, up, right...".
Why not write those thoughts down? Using "u" and "r" we can write out a path:
r r r r r r u u u u
That is, go all the way right (6 r's), then all the way up (4 r's).
The path in the diagram would be:
r r r r u u u u r r
Using the text interpretation, the question becomes "How many ways can we re-arrange the letters rrrrrruuuu?"
Ah, the ubiquitous combination/permutation problem -- never thought it'd be useful, eh?
Understanding Combinations And Permutations
There's several ways to see combination and permutation problems. Once the first explanation clicks, we can go back and see it a different way. When trying to build math intuition for a problem, I imagine several mental models circling a core idea. Starting with one insight, I work around to the others.
Approach 1: Start The Same
Instead of having 6 rights at 4 ups, imagine we start with 10 rights (r r r r r r r r r r).
Clearly this won't do: we need to change 4 of those rights into ups. How many ways can we pick 4 rights to change?
Well, we have 10 choices for the first 'right' to convert (see the combinations article). And 9 for the second, 8 for the third, and 7 choices for the final right-to-up conversion. There are 10 * 9 * 8 * 7 = 10!/6! = 5040 possibilities.
But, wait! We need to remove the redundancies: after all, converting moves #1 #2 #3 and #4 (in that order) is the same as converting #4 #3 #2 #1. We have 4! (4 * 3 * 2 * 1 = 24) ways to rearrange the ups we picked, so we finally get:
We're just picking the items to convert (10!/6!) and dividing out the redundancies (4!).
Approach 2: Just Use the Combination Formula
Halfway through that explanation, you might have realized we were recreating the combination formula:
That's the shortcut when you know order doesn't matter. However, sometimes I'm not sure whether I need a permutation or combination from the outset. While saying "Just use C(10,4)" may be accurate, it's not helpful as a teaching tool. Sometimes it helps to re-create the situation on your own.
Approach 3: Start Different
Here's another approach: instead of letting each r and u be interchangeable, label the 'right' moves r1 to r6, and the 'up' moves u1 to u4. How many ways can we re-arrange these 10 items?
This question is easy: 10! = 3,628,800 (wow, big number). We have 10 choices for the 1st move, 9 for the second, and so on, until we have 2 choices for the 9th and only 1 for the last. Cool.
Of course, we know that "r1 r2 u1 u2" is the same path as "r2 r1 u2 u1". We can shuffle the r's and u's in their own subgroups and the path will stay the same.
• How many ways can we shuffle all 10? 10! = 3,628,800
• How many ways can we shuffle 6 r's? 6! = 720
• How many ways can we shuffle 4 u's? 4! = 24
So, we start with the total number of possibilities (10! = 3,628,800) and divide out the cases where we shuffle the r's (6! = 720) and the u's (4! = 24):
Neat! It's cool seeing the same set of multiplications and divisions in different ways, just by regrouping them.
Definition:
Permutation:
An arrangement is called a Permutation. It is the rearrangement of objects or symbols into distinguishable sequences. When we set things in order, we say we have made an arrangement. When we change the order, we say we have changed the arrangement. So each of the arrangement that can be made by taking some or all of a number of things is known as Permutation.
Combination:
A Combination is a selection of some or all of a number of different objects. It is an un-ordered collection of unique sizes. In a permutation the order of occurence of the objects or the arrangement is important but in combination the order of occurence of the objects is not important.
Formula:
Permutation = nPr = n! / (n-r)!
Combination = nCr = nPr / r!
where,
n, r are non negative integers and r<=n.
r is the size of each permutation.
n is the size of the set from which elements are permuted.
! is the factorial operator.
Example:Find the number of permutations and combinations: n=6; r=4.
Step 1: Find the factorial of 6.
6! = 6×5×4×3×2×1 = 720
Step 2: Find the factorial of 6-4.
(6-4)! = 2! = 2
Step 3: Divide 720 by 2.
Permutation = 720/2 = 360
Step 4: Find the factorial of 4.
4! = 4×3×2×1 = 24
Step 5ivide 360 by 24.
Combination = 360/24 = 15
The above example will help you to find the Permutation and Combination manually.
1. Determine whether the following situations would require calculating a permutation or a combination:
a.) Selecting three students to attend a conference in Washington, D.C.
permutation combination
b.) Selecting a lead and an understudy for a school play.
permutation combination
c.) Assigning students to their seats on the first day of school.
permutation combination
2. A teacher is making a multiple choice quiz. She wants to give each student the same questions, but have each student's questions appear in a different order. If there are twenty-seven students in the class, what is the least number of questions the quiz must contain?
3. A coach must choose five starters from a team of 12 players. How many different ways can the coach choose the starters?
4. The local Family Restaurant has a daily breakfast special in which the customer may choose one item from each of the following groups:
Breakfast Sandwich Accompaniments Juice
egg and ham
egg and bacon
egg and cheese breakfast potatoes
apple slices
fresh fruit cup
pastry orange
cranberry
tomato
apple
grape
a.) How many different breakfast specials are possible?
b.) How many different breakfast specials without meat are possible?
5. In how many ways can 3 different vases be arranged on a tray?
6. There are fourteen juniors and twenty-three seniors in the Service Club. The club is to send four representatives to the State Conference.
a.) How many different ways are there to select a group of four students to attend the conference?
b.) If the members of the club decide to send two juniors and two seniors, how many different groupings are possible?
排列（有顺序）：mPn＝m*（m-1）*.....*(m-n+1)
组合（无顺序）:mCn＝m*（m-1）*.....*(m-n+1)/(1*2*...*n)
排列：从n个不同的元素中取m(m≤n)个元素，按照一定的顺序排成一排，叫做从n个不同的元素中取m个元素的排列。
排列数：从n个不同的元素中取m(m≤n)个元素的所有排列的个数，叫做从n个不同元素中取出m个元素的排列数，记为Pnm
排列公式：p(n,m)=n*(n-1)*.....(n-m+1)
组合：从n个不同的元素中，任取m(m≤n)个元素并成一组，叫做从n个不同的元素中取m个元素的组合。
组合数：从n个不同的元素中取m(m≤n)个元素的所有组合的个数，叫做从n个不同元素中取出m个元素的组合数，记为Cnm
组合公式：c(n,m)=p(n,m)/m!=n!/(m!*(n-m)!)
2. If there were two questions on the quiz, we could prepare two quizzes with the questions in different order -- 2•1 = 2.
If there were three questions, we could get 3•2•1 = 6 different orders.
If there were four questions, we could get 4•3•2•1 = 24 different orders -- not quite enough for the class of 27 students.
If there were five questions, we could get 5•4•3•2•1 = 120 different orders. The teacher will need at least 5 questions on the quiz.
3.
Choose 5 starters from a team of 12 players. Order is not important.
4.
a.) Basic counting principle:
Sandwiches x Accompaniments x Juice
3 • 4 • 5 = 60 breakfast choices
b.) Meatless means that under Sandwiches there will be only one choice.
Sandwiches x Accompaniments x Juice
1 • 4 • 5 = 20 meatless breakfast choices
5. Basic counting principle:
3! = 3•2•1 = 6 ways
6. 14 juniors, 23 seniors
37 students total
a.) Choose 4 students from the total number of students. Order is not important.
b.) Choose 2 juniors and 2 seniors.
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我没看过icas的题，不知道啊~~~我乱说的哈，我觉得selective难吧，小胖子icas从来不准备的，似乎拿distinction挺容易的，selective就很不容易了，错一道题，排名就靠后好多
朱兄的方法更容易理解，但是真到考试的时候，你只有一分钟来做出这道题，我个人觉得还是记住那两个公式，只要理解该用那个公式，一下子就能算出来.....不好意思，反对"应试"的同学不要砸我哈~~~
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这是我那个文件里的，文件里还有图，更容易理解些。俺再贴一遍pfd的吧。
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奇怪我是去年一个山区朋友给的，难到你是？
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