How many numbers with distinct digits are there between 1000 and 9999. [1]
I came up with a solution like this.
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….
[1] Exercise 2.7.15 from Applied Combinatorics 2nd edition by Fred S. Roberts and Barry Tesman
Best Answer
Your problem comes when you assume that there were $3$ taken from your last set. It's quite possible that the second or third choice was a $0$, so that you in fact have only eliminated $2$ options from the last set.
Instead, choose the last digit first ($5$ choices), then the first digit ($8$ choices, since we can't have picked $0$ for the last digit), then deal with the second and third digit ($8$ and $7$ choices, in some order). That will get you $5\cdot 8\cdot 8\cdot 7=2240$, as desired.