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  • Count the total number of different six-digit numbers that can be created using the digits 0, 1, 2, 3, 4, and 5, allowing repetition, where the sum of all digits is even<

Count the total number of different six-digit numbers that can be created using the digits 0, 1, 2, 3, 4, and 5, allowing repetition, where the sum of all digits is even

 

Option: 1

2917
 


Option: 2

2685
 


Option: 3

3241
 


Option: 4

2404


Answers (1)

best_answer

To count the total number of different six-digit numbers that can be created using the digits 0,1,2,3, 4 , and 5 , allowing repetition, where the sum of all digits is even, we can consider the different cases.

Case 1: Even number of odd digits ( 0 or 2 or 4 or 6 ):

In this case, there are three even digits (0,2, and 4) and three odd digits (1,3, and 5). We need to choose the positions of the even digits and odd digits in the six-digit number. There are several sub-cases:

Sub-case 1: All even digits:

There is only one possibility in this sub-case, which is using all even digits.

Sub-case 2: Two even digits and four odd digits:

There are six choices for the positions of the even digits and three choices for each even digit. The remaining positions can be filled with any of the odd digits. Therefore, there are 6 \times 3 \times 3 \times 3 \times 3 \times 3=1458 possible arrangements in this sub-case.

Sub-case 3: Four even digits and two odd digits:

There are six choices for the positions of the odd digits and three choices for each odd digit. The remaining positions can be filled with any of the even digits. Therefore, there are 6 \times 3 \times 3 \times 3 \times 3 \times 3=1458 possible arrangements in this sub-case.

Case 2: Odd number of odd digits ( 1 or 3 or 5 ):

In this case, there are three even digits (0,2, and 4) and three odd digits (1,3, and 5). We need to choose the positions of the even digits and odd digits in the six-digit number. There is only one sub-case:

Sub-case: One even digit and five odd digits:

There are six choices for the position of the even digit and three choices for the even digit. The remaining positions can be filled with any of the odd digits. Therefore, there are 6 \times 3 \times 3 \times 3 \times 3 \times 3=1458 possible arrangements in this sub-case.

Therefore, the total number of different six-digit numbers that can be created, where the sum of all digits is even, is 1 (from the first case, sub-case 1) +1458 (from the first case, sub-case 2 and sub-case 3) +1458 (from the second case, sub-case) =2917.

It is important to note that repetition is allowed in this case, as the digits 0,1,2,3,4, and 5 can be used multiple times.

Posted by

Ritika Jonwal

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