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  • If the digits 0, 1, 2, 3, 4, and 5 are arranged to form a six-digit number, how many different numbers can be created, where the third digit must be even? &nbsp

If the digits 0, 1, 2, 3, 4, and 5 are arranged to form a six-digit number, how many different numbers can be created, where the third digit must be even?

 

Option: 1

2500

 


Option: 2

6000

 


Option: 3

1440

 


Option: 4

2450


Answers (1)

best_answer

To find the number of different six-digit numbers that can be created using the digits 0,1,2,3,4, and 5 , where the third digit must be even, we need to consider the possibilities for the positions of the digits.

Since the third digit must be even, there are 3 choices for the third digit: 0,2 , or 4 . After choosing the third digit, there are 5 remaining digits to choose from for the first position, 4 choices for the second position, 4 choices for the fourth position, 3 choices for the fifth position, and 2 choices for the sixth position.

Therefore, the total number of different six-digit numbers that can be created, where the third digit is even, is

3 \times 5 \times 4 \times 4 \times 3 \times 2=1,440.

It is important to note that repetition is not allowed in this case, as the digits 0,1,2,3,4, and 5 are distinct.

Posted by

Pankaj Sanodiya

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