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  • In how many ways can the digits 0, 1, 2, 3, 4, and 5 be arranged to form a six-digit number, where the last digit must be even?  <div class='qna-option

In how many ways can the digits 0, 1, 2, 3, 4, and 5 be arranged to form a six-digit number, where the last digit must be even?

 

Option: 1

360


Option: 2

520


Option: 3

460


Option: 4

240


Answers (1)

best_answer

To find the number of ways the digits 0,1,2,3,4, and 5 can be arranged to form a six-digit number, where the last digit must be even, we need to consider the possibilities for the positions of the digits.

Since the last digit must be even, there are 3 choices for the last digit: 0,2 , or 4 . After choosing the last digit, there are 5 remaining digits to choose from for the first position, 4 choices for the second position, 3 choices for the third position, 2 choices for the fourth position, and 1 choice for the fifth position.

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

3 \times 5 \times 4 \times 3 \times 2 \times 1=360.

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

HARSH KANKARIA

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