Question

How many odd numbers between 1000 and 9999 have distinct digits?

Answer 1

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Since we can't know what numbers have been used, in the tens, hundreds and thousands we start counting at the ones.

1s: { 1, 3, 5, 7, 9 }, so 5 initial possibilities

10s: { 0, 1, ... , 9 }, so 10 initial possibilities, 1 taken: 9 left

100s: { 0, 1, ... , 9 }, so 10 initial possibilities, 2 taken: 8 left

1000s: { 1, 2, ... , 9 }, so 9 initial possibilites, 3 taken: 6 left

So then we arrive at the following: 5 * 9 * 8 * 6 = 2160 possibilities. I thought this was pretty straight forward.

Than I had a glimpse at the solution sheet... And lo an answer which really doesn't make much sense at its first glimpse.

Calculate the sum of those odd numbers with distinct digits with no 0’s, a 0 in the tens place, or a 0 in the hundreds place. No 0’s: 5 choices for the ones place, then 8 · 7 · 6 choices for the other three places; 0 in the tens place: 5 choices for the ones place and 1 choice for the tens place, then 8 · 7 choices for the other two places; 0 in the hundreds place: 5 choices for the ones place and 1 choice for the hundreds place, then 8 · 7 choices for the other two places;

(5 · 8 · 7 · 6) + (5 · 1 · 8 · 7) + (5 · 1 · 8 · 7) = 2240;

Why are the 0's treated special? The exercise states it should be an odd number, with distinct digits. I thought I adhered to that proposition....

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