Place 10 books on any row of shelves and find the probability of three of the specified books being

The probability of putting together the three books specified: 1 15.

First of all, arrange the specified three books, there are a total of A3 3 arrangement, that is, 6 types.

Then arrange the other seven books, a total of A7 7 arrangement, that is, 5020 kinds.

The three designated books are inserted as a whole by interpolation, so there are a total of 8*6*5020 ways to arrange them.

Arbitrarily arrange these ten books, there are a total of A10 10 10 ways to arrange.

So the probability is 8*(a3 3)*(a7 7) (a10 10)=1 15.

Put 10 books on the shelf, that is, make 10 books in full row, a total of 10! kind, and then take out the designated three books for a full row of 3! Kind, and then the three designated books are considered as a whole and the other 7 books are all arranged with 8!

species, according to the principle of step-by-step calculation, three designated books are arranged together for a total of 3! x8！；So the probability is:

3！Multiply by 8! Divide with 10!

The answer is 1 15

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Problem description: Place 10 books on a random row of bookshelf, and find the probability of putting the specified 3 books together. Please write the steps and results.

Analysis: Solution: First of all, regard the three books as an element, and arrange all 8 elements with the other seven books, and then reverse the positions of the three books.

So the result is 8!*3!/10!=1/15

Specify the probability of 3 books being cracked up.

- Method 1: p(8,8)*p(3,3) skillfully p(10,10)=3*2*1 (10*9)=1 15

- Method 2: c(8,1) Filial piety and pure c(10,3)=8*1*2*3 (10*9*8)=1 15

Full permutation a(10,10) = 10!

Arbitrary position: c(8,1) Other books: any arrangement: a(7,7), three Kai debates, this book, any arrangement, Sun Zhilie, grip elimination a(3,3).

p=8×6/10/9/8=1/15

Problem solving ideas: This problem is the probability of a possible event, the event contained in the occurrence of the experiment is 10 books randomly put together, a total of a 10 10 results, the event that satisfies the condition is the specified 3 books put together, that is, the three books are regarded as an element, and the other 7 books are arranged, and there is an arrangement between the three books to obtain the probability

From the meaning of the question to know the probability that this question is an equal possible event, the event contained in the occurrence of the test is 10 books casually placed in a balance line, a total of a1010 results, the event that satisfies the conditions is the specified 3 books put together, that is, the three books are regarded as an element, and the other 7 books are arranged, and there is a town arrangement between the three books, a total of a88a33, in which the probability of the specified 3 books put together is.

a88a33

a1010=[1 15], so the answer should be cautious: [1 15]7, the three books together are recorded as a whole a, and there are a total of 3!A and all possible permutations of the other 7 books are 8!Seed.

There are 10 out of 10 books in all the arrangements!Seed.

Therefore, the probability is 3!8!/10!=1/15。, 1, Combination Question 10! /c（10，3）

Here's the answer, 0,

10 books are randomly arranged on the book bush digging shelf, and there is a total of a(10,10)=10!methods.

The method of putting together the specified 5 books is a(5,5)*a(6,6)=5!*6!methods.

So, the probability is 5!*6!/10!=5!/(10*9*8*7)=120/(720*7)=1/42

The remaining seven books are not considered, and the ABC of the 3 books is placed exactly in the 123 position of 10 positions, with a probability of 1 (10*9*8).

There are six ways in which these three books are arranged on their own.

Therefore, regardless of the order in which the three books are placed in the first three positions, the probability is 6 (10*9*8).

There are 8 ways to put three books in 10 positions together.

The probability of three books being put together is the sum of all ways (6*8) (10*9*8)=1 15

The probability of the specified 3 books being put together is p(8,8)*p(3,3) p(10,10)=3*2*1 (10*9)=1 15

Look at the 3 books as a whole, and there are 7 books left, and think about this question in the blanks.

7 books, with a total of 8 empty slots, can be calculated as follows:

The 10 books are arranged in a random arrangement.

Put the 3 books specified in it together and make 1 large book and the remaining 7 books randomly arranged c (the arrangement of these 3 books c (

The probability is c (

The A and C of the recommended answer are mistaken.