This is a combinatorics problem I must solve:
How many even numbers between 10,000 and 20,000 have distinct
digits(i.e no repetitions)?
Is the way I solved it correct?
I put for lines for the numbers :
_ , _, _ ,_ , _
And I put numbers in each spot: 1, 8, 7, 6, 4
Multiplying out I get 1,324 . Is this the right way to go?
Best Answer
Almost. You’re correct that there is just one possibility for the first digit. Now start at the other end: an even number must end in $0,2,4,6$, or $8$, so there are five possible last digits. Once the leading $1$ and the even last digit have been chosen, there are $8\cdot7\cdot6$ possibilities for the middle three digits, for a total of $$1\cdot5\cdot8\cdot7\cdot6=1680\;.$$ (The other problem with your answer is that you multiplied incorrectly: you should have got $1344$, not $1324$.)