Question 23

The number of 5-digit numbers consisting of distinct digits that can be formed such that only odd digits occur at odd places is

We need to form a 5-digit number.

Assume abcde is that 5-digit number.

The positions will be 1st, 2nd, 3rd, 4th, 5th for a, b, c, d and e respectively.

Odd places: 1st, 3rd, 5th

Even places: 2nd, 4th

The Digits available are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

We need to use distinct digits (no repetition) and use only odd digits (1, 3, 5, 7, 9) at odd positions

Odd positions are 1st, 3rd, 5th place.

Available odd digits = {1, 3, 5, 7, 9} = 5 digits

We have to fill three places with these 5 digits without repetition.

1st place (For a) = 5 options

3rd place (For c) = 4 options

5th place (For e) = 3 options

Number of ways = 5 × 4 × 3 = 60

Even positions are 2nd and 4th place.

Available digits now = total 10 digits − 3 already used = 7 remaining digits

2nd place (For b) = 7 options

4th place (For d) = 6 options

Number of ways = 7 × 6 = 42

Total numbers = (Ways to fill odd places) × (Ways to fill even places)

=> 60 × 42

= > 2520

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