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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 ...
Understanding the problem:
To form a 7-character code with the given conditions, we need to place even digits at even positions and odd digits at odd positions. We can use the digits 1-9 with repetition allowed.

Solution:
- Determine the number of ways to choose even digits:
- There are 5 even digits (2, 4, 6, 8) available.
- The even positions in the code are 2nd, 4th, and 6th.
- Therefore, the number of ways to choose even digits = 5 * 5 * 5 = (5)^3
- Determine the number of ways to choose odd digits:
- There are 4 odd digits (1, 3, 5, 7, 9) available.
- The odd positions in the code are 1st, 3rd, 5th, and 7th.
- Therefore, the number of ways to choose odd digits = 4 * 4 * 4 * 4 = (4)^4
- Calculate the total number of codes:
- To find the total number of codes, we multiply the number of ways to choose even digits and odd digits.
- Total number of codes = (5)^3 * (4)^4 = (5)^4 * (4)^3
Therefore, the correct answer is option (C) (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?
Question Description
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