Broad combination (learning notes)

[Theorem] X is a set of k elements. It performs n copies of the elements in X in non-sequential order.

Possible results of C (n + k-1, k-1.

Note that it is a set and the elements are differentiated.

The alternative is not so abstract: the number of possible results for a ball with K colors. N balls are taken from the ball.

(Each color ball is not differentiated, with a large number: n> N for any N, each color ball)

The teaching material may have such a difficult name called "partition method": Put a K-1 in N + k-1 stool "partition"

(K-1 partition can divide the remaining n stools into K classes, each class corresponds to a color)

Medium K = 3, n = 6; green corresponds to the partition, yellow corresponds to the ball ..

[Example]

There are a bunch of red balls, a bunch of blue balls, and a bunch of green balls, each with no less than 8.
(1) How many different pumping methods are there to extract 8 balls from them?
(2) How many different pumping methods are required for every 8 balls in each color?
[Solution] (1) k = 3, n = 8, so the number of solutions C (8 + 3-1-1) = 45
(2) Take one ball of each color first and then convert it
(1) In this case, K = 3, n = 5, so the number of solutions C (5 + 3-1-1) = 21

[Example] the equation in the integer field is X1 + X2 + X3 + X4 = 29. Q:
(1) how many non-negative integer solutions does the equation have?
(2) How many of the equations meet the requirements of x1> 0, X2> 1, X3> 2, X4

> = Integer solution of 0?

[Solution] (1) the solution for each non-negative integer of the equation is equivalent to copying 29 elements from four elements. The number of elements copied from the I element is Xi, I = 1 .. 4. Therefore, the number of solutions C (29 + 4-4960-1) =
(2) select at least one of the 1st elements, at least two of the 2nd elements, at least three of the 3rd elements, and the remaining 23 can be selected as needed. Therefore, the number of solutions is C (23 + 4-2600-1) =.