12.AB must be adjacent, AB can be packaged as a person as G, and the topic is translated into, CDEFG5 objects line up in a row, a * * *, how many kinds of stations are there? This is very simple, A55 (explain a little why A55, five people are in a row, a * * has five positions, and the first position is one of five people. . . The reverse is 5*4*3*2* 1, which is A55). Because the title only says that AB is adjacent, AB and BA are two different situations in the process of AB packaging. So, the final result is A55*2.

Abs cannot be adjacent. Change your mind. The four children corresponding to CDEF4 are arranged in a row first (so that there are five positions in front of C, between CD, DE and EF, and behind F, there is a * * *), and then the AB children are not adjacent. So the result is A44*A52(A44 is the combination number of four little friends standing in a row, A52 is the combination number of five children inserting AB).

13. (1) * * How many drawing methods are there? 10 Choose 3, because there is no order, so it is C 10 3.

(2) There is only one defective product. Analysis, draw 3, just one defective product, and the remaining 2 are genuine. The problem is to take one out of three defective products and two out of seven good products. C3 1*C72

(3) There is at least one defective product. Two ways of thinking.

One is that there is at least one defective product, and the number of defective products can be 1, 2, 3. This is equivalent to three questions 1. Exactly 1 defective product, exactly 2 defective products and exactly 3 defective products. The combination numbers are C3 1 * C72, C32 * C7 1 and C33 * C70 respectively. Add them together.

The second is that there is at least one defective product+no defective product = all drawing methods.

None of the defective products is C73, and all drawing methods are C 10 3. Then at least one defective product is C 10 3-C73.

14. This can also be converted into extracting four numbers from the number of 19, the largest one is a single digit, the second largest is a thousand digits, the third largest is a hundred digits, and the smallest is a ten digits. So the result is C 10 4.

15. Take at least one pill a day. Under the transformation of the problem, 10 sugar is arranged in a row, and two knives are cut from the middle. Emm is two knives and three pieces, and one piece is eaten every day. 10 candy, there are 9 spaces for you to have a knife. The answer is C92.

Take at least 2 capsules a day. Similarly, turn three of the candies into super candies, that is, one candy is two candies, so that there are seven candies left, arranged in a row, cut two knives in the middle and eat one every day. Seven sweets and six spaces can give you a knife. The answer is C62.

In other words, it is 10 candy. Take out three sweets first and eat one every day. The problem becomes, 7 pieces of sugar, eat at least one piece a day. O(∩_∩)O haha ~

One tablet a day is allowed.

This question is equivalent to: 13 sweets, eat at least one candy every day, and how many ways to eat a candy * * * is obviously the same as the question 1. Choose two sweets from 12 spaces. C 12 2。