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  • In how many ways can the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 be arranged to form a nine-digit number by swapping the positions of any two digits?  <di

In how many ways can the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 be arranged to form a nine-digit number by swapping the positions of any two digits?

 

Option: 1

253,745

 


Option: 2

789,526

 


Option: 3

362,880

 


Option: 4

241,000


Answers (1)

best_answer

To find the number of ways the digits 1,2,3,4,5,6,7,8, and 9 can be arranged to form a nine-digit number by swapping the positions of any two digits, we need to consider the different arrangements that result from swapping the positions of any two digits.

Since there are nine digits, there are a total of 9 choices for the first digit. After choosing the first digit, there are 8 remaining digits to choose from for the second digit. Similarly, there are 7 choices for the third digit, 6 choices for the fourth digit, and so on, until there is 1 choice for the last digit.

Therefore, the total number of different nine-digit numbers that can be formed is

9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=362,880.

It is important to note that swapping the positions of two digits does not affect the number of different numbers that can be formed in this case, as all nine digits are distinct.

Posted by

Shailly goel

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