Math Combination and Permutation One Problem, How to Do This Math Problem Permutation and Combinati

At the head of the first row, there are a total of 4a4 = 24 types of arrangement.

At the end of the second row, there are also 24 kinds.

If five people are asked to do a full arrangement, there are 5a5 = 120 species.

But A and B at the same time at the beginning and end of the situation counted one more time, that is, 3A3 = 6 kinds.

Subtracting the impermissible from all the situations is what is desired.

120-24-24+6=78 species.

If you don't count the permutations of others.

then 6 kinds. A and B??

B A??? A and B??

B A??? Armor? Second?

Second? Armor?

Solution: According to the title, the arrangement of A and B has the following situations:

1, A and B

2, A and B

3, A and B

4, B A

5, B A

6, B A

When the A and B station method is fixed, the other three people have: A3 = 3 * 2 * 1 = 6 kinds of arrangement;

When there are six cases of the position of A and B, there are a total of 6*6=36 types.

A: There are 36 different standing methods.

This is 5 people, because A and B do not occupy the first and last rows, so they have 3*2 standing methods, and the rest of the people are 3*2*1

So the total is 3*2*3*2*1=36

Mainly use the plug-in method:

x+y+z=8, find the number of positive integer ungroups of xyz.

Answer: c(7, 2) = 21

The equivalent of eight balls is lined up in a row and divided into three piles. Insert two boards in the gap between the balls, not at both ends, so that they are divided into three groups, and the number of balls in each group corresponds to the solution of xyz.

Find, non-negative integer solution:

Answer: c(10, 2) = 45

Unlike the first question, xyz can take a zero value, and many students will write c(9, 2) as a matter of course, thinking that the endpoint is enough. However, it should be taken into account that xyz can be taken as zero, and c(9, 2) ignores x=0, y=0, and c=0. So let's do a one-step transformation:

x+y+z=8 ⇔ x+1) +y+1) +z+1) =11

This brings us back to the first question, the sum of three positive integers to 11.

Find the solution of a non-negative integer for x+y+z <=8:

Answer: c(11, 3) = 165

This question is difficult, but with the foreshadowing of the first two questions, it is very simple.

Obviously, when xyz satisfies the problem, there is 8-x-y-z, which is a non-negative integer. ∴ x+y+z+ξ 8

This translates to the first problem, which is to find the number of groups of four numbers and the solution of a non-negative integer with 8.

If the two left and right hands of a pair of gloves are exactly the same, then there are c(11,8)=165 ways.

If a pair of gloves is divided into different left and right hands, there are c(11,8)*2 8=42240 ways to take it.