A type of knowledge about the number of combinations of fact, the key is how to count so many problems to solve, the problem here is counting method commonly used model partition.
Specific method refers to the separator n digital program into a number of groups m counting it seems difficult to have an idea is that each number has a selection answer m ^ m n But there is divided by repeated n! There is no specific meaning it can not do so.
We observed that these balls are not different from each other. It is known not pay attention to the number of combinations different from each other so consider switching to a combination of the model number. We know that the number m in the group numbers add up to n.
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Abstract model n balls are putting in place a row of plates m-1 can not pay attention to this issue is completed mutually different. However, we consider only the discharge plate so is C (n-1, m-1) in the void.
However, we consider each component to the number of balls is> = 1 so that when the topic that may be 0 when we need to do an operation in each case we put a ball as the ball so put 0 we just need to turn decreasing more out n the ball just fine, thus ensuring that each group have at least had a ball, of course, if it is a ball, then the ball is actually 0.
Here it leads to the definition of our separator law: n-1 empty plate which is inserted into the k k + 1 group in the method of the n elements.
Three necessary conditions for the application of the separator Method d: 1 n mutually different elements must not be divided by each group 2 must have an element of group 3 is divided into different from each other.
Next comes several variations of it ...
Tim separator element method. Application of the above-described embodiment is actually a primer such a method.
1 then replace the same question 10 small balls into three different boxes on a box at least a second case at least three months a third box can hold the number of cases ... it is also simple .. .
Of course, because the ball is the same we can directly put a King James the first ball put three balls in a box to put the second box then 7 three balls into the box and each box can hold the ball and we the successful conversion of the above model ...
But this is still around in a circle because we ultimately converted into positive integer solution, it is better to convert consider our handpicked a box we put two last manually add a second ball from the box positive integer solution to allow direct representation of third a box with at least one ball when the ball is represented as 0 2 represents a ball when the ball ... can be.
Why proved correct ... it is the same one found the right formula to both approaches, we prove that 2 is always defined the third ball of the box, then that is the number of balls this could prove to be correct -1.
2 the same 20 numbered balls placed in each of four cages 1,2,3,4 required number of balls in each box number is not less than its number, total number of seek discharge method.
Obviously we direct Imperial put on the line anyway, the ball is the same. Proof of correctness can be proved from the existence of the answer.
3 a class of natural numbers beginning of each number from the third number is equal to the sum of its first two digits up until the request can not write this kind of digital how many?
Obviously we've never 10-100 burst start program number search can search directly, but we can use the above method to solve this problem partition it? I just figured I find the manual is 45 this also means that no use for, but I want us to partition method to solve this problem ...
Obviously a digital fixed then the whole numbers we just need to fix the first two to statistics from the first two legitimate. The first set was a second place there is b then a + b <= 9 and a> 0 this time we put <= == is replaced by a top model ... unfortunately not.
We found that b can be zero so the first meeting to let the ball b> 0 is a + b <= 10 ab are considering is a positive integer <how to deal with this is not necessarily all 10 balls into a and b and every time we did not put ab put the ball into the c enough then there is a + b + c == 10 but there may be done and c is empty again so we handpicked c> 0 then the result is
11 among the three balls into the box C (10,2) == 45; very clever use of the principle of continued conversion.
Tim flapper plate method. Or the above examples, we make specific this is to add a c flapper plate and then the above process is this, however, and add elements of the collection together.
Method palette. There are 10 sugar, if you eat every day at least one (and more limited), finished up, ask how many different eat? We seem difficult to solve the problem using the palette method.
First, we have the necessary if the plates 9 are put 10 days, then a daily described, each plate may choose to put a sugar or hold so defined between the two plates ^ 9 = 512 as a front plate that a day to eat. This perfect conversion of the problem and does not weight not leak.
Of course there is a proof that we first put up the nine board found that if one day I eat a day explained last day is not ... that is the last day to eat nine kinds penultimate day is eight kinds .. doing so obviously wrong because there are duplicate the ...
But in fact, we draw a picture look at the above approach is clearly correct.
Classification flapper law. Xiaomei has 15 pieces of candy, if you eat at least three, finished up, then eat a total of how many different?
The above method of palette is very similar, but we can not guarantee every day to eat three direct Imperial does not work we do not know the number of days not Imperial ...
We can talk about the classification of eating up to five days so most eat one day these two programs are 1, when to eat two days, when King James ... C (10,1) = 10 eat for three days when the Imperial ... C (8,2) = 28. 4 days to eat by Imperial ..C (6,3) = 20
There are 60 kinds of the ensemble.
Gradually flapper law. The original six programs in a single program, if these programs remain relatively the same order, and then add three programs, there are several situations?
We can gradually inserted into the insertion of a plate when the gap 7 C (7,1) the second continued C (8,1) a third C (9,1) obtained by multiplying 504; because we add three mutually different elements can not be inserted together to progressive insertion.
Basically all the models here bulkhead law ended when the counting method to memorize the number of good / cy