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A 7-character code is such that only even digits occur at even places and only odd digits occur at odd places, e.g. 1436789. How many such codes can be made from digits 1-9, if repetition of digits is allowed?
  • a)
    (5)4 × (5)3
  • b)
    (4)4 × (5)3
  • c)
    (5)4 × (4)3
  • d)
    (5)4 × (4)5
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A 7-character code is such that only even digits occur at even places...
Solution:

Given that 7-character code is such that only even digits occur at even places and only odd digits occur at odd places.

We need to find the number of such codes that can be made from digits 1-9, if repetition of digits is allowed.

Let us consider the even places first.

The even places can be filled with even digits 2, 4, 6, and 8. So, we have 4 options for each even place.

The odd places can be filled with odd digits 1, 3, 5, 7, and 9. So, we have 5 options for each odd place.

Therefore, the total number of possible codes is:

4 × 5 × 4 × 5 × 4 × 5 × 4 = (5)4 × (4)3

Hence, the correct answer is option C.
Free Test
Community Answer
A 7-character code is such that only even digits occur at even places...
Odd digits: 1, 3, 5, 7, and 9
Even digits: 2, 4, 6, and 8
Number of ways to fill an odd place = 5
Number of ways to fill an even place = 4
Therefore,
Total number of cases = (5)4 × (4)3
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A 7-character code is such that only even digits occur at even places and only odd digits occur at odd places, e.g. 1436789. How many such codes can be made from digits 1-9, if repetition of digits is allowed?a)(5)4 × (5)3b)(4)4 × (5)3c)(5)4 × (4)3d)(5)4 × (4)5Correct answer is option 'C'. Can you explain this answer?
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