Paste first reference blog (1) (2) (3) // have quality
Method baffle element that is inserted into the n (n-1 empty) of the k-1 board, put into k groups of n elements, the program number $ C_ {n-1} ^ {k-1} $
Example 1:
The 10 identical small balls into 3 boxes, each box at least one , and asked several situations?
Or find the equation x + y + z = positive integer number 10 Solution
Answer: $ C_ {10-1} ^ {3-1} = $ $} ^ {C_ {2}. 9 $
Example 2:
The 10 identical small balls into 3 boxes, each box can hold , and asked several situations?
Or find the equation x + y + Solution nonnegative integer number of z = 10
We assume that for each bin of each add a ball , the problem is converted to the same problems as in Example 1:
The 13 identical balls into three different boxes, each box at least one, there are several situations?
Answer: $ 12 is C_ {$} ^ {2}
Well, that is entered :( The Thought)
Tim baffle element method
Example 3:
The 10 identical balls placed . 3 different boxes, the first boxes at least . 1 a , the second case at least . 3 months , the third box can hold the ball, there are several situations?
Give the third boxes add a ball , and then out from two balls 10 into a first box, and the problem Conversion Example 1:
The 9 (10 + 1-1) and a small ball into three different boxes, each box at least one, several situations?
Answer: $ C_. 8} {2} {$ ^
Example 4:
The 20 same pellet into four boxes were numbered 1, 2, the required number of balls in each box number is not less than its number , total number of seek discharge method. (Reducing the number of spacer balls method)
First out of the six balls 20 are divided into two, three, four boxes were placed in 2,3,4, the problem becomes (again Example 1):
The 14 identical balls into four different boxes, each box at least one, there are several situations?
Answer: $ 13 is C_ {} ^ {} $. 3
Example 5:
There is a class of natural numbers, from the third number begins each number it happens to be in front of two numbers and , until not write so far, such as 257,1459 and so on, there are a few such numbers?
Properties: (1) determining a number of the first two (2) provided for the first two a, b, then a + b <= 9, a is not 0 and
So just looking to meet the first two (2) There are several cases
Then I wrong into the problem: the 9 divided into two parts, the first part> = 1, the portion may be 0 ==> 9 balls into the two boxes, one of at least a first, second can hold ==> A case 3, the answer is $ C_ {9} ^ {1 } $ think about this question and Example 3 What is the difference? Ah, Example 3 10 balls must all placed, and this problem is < = 9
So how do?
1 idea: we add a
Tim baffle plate method