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  • How many integers can be formed using the digits 3, 5, 6, 7, and 8, with repetition, that lie between 6000 and 7000? Additionally, each integer must contain at least one odd digit and one even digi

How many integers can be formed using the digits 3, 5, 6, 7, and 8, with repetition, that lie between 6000 and 7000? Additionally, each integer must contain at least one odd digit and one even digit.

 

Option: 1

555


Option: 2

785


Option: 3

465


Option: 4

625


Answers (1)

best_answer

To find the number of integers that can be formed using the digits 3, 5, 6, 7, and 8 with repetition, and lie between 6000 and 7000 while also ensuring that each integer contains at least one odd digit and one even digit, we can consider the following cases:

Case 1: The first digit is odd (3, 5, or 7)

In this case, we have 3 choices for the first digit. For the remaining three digits, we have 5 choices for each digit since repetition is allowed. So, the total number of integers in this case is:
3 \times 5 \times 5 \times 5=375
 

Case 2: The first digit is even (6 or 8)

In this case, we have 2 choices for the first digit. For the remaining three digits, we have 5 choices for each digit. So, the total number of integers in this case is:
2 \times 5 \times 5 \times 5=250

Adding up the two cases, we get:
375+250=625
Therefore, the number of integers that can be formed using the digits 3, 5, 6, 7, and 8 with repetition, and lie between 6000 and 7000 while also ensuring that each integer contains at least one odd digit and one even digit is 625 .

 

Posted by

Ritika Kankaria

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