- Background Information:
I am studying discrete mathematics, as I was practicing, I came across this problem and solution in my textbook. However, I cannot fully understand the solution. I need some clarification, thanks.
Determine the number of six-digit integers (no leading zeros) in which (a) no digit may be repeated; (b) digits may be repeated. Answer parts (a) and (b) with the extra condition that the six-digit integer is (i) even; (ii) divisible by 5; (iii) divisible by 4.
- Textbook Solution:
For part b i)
ANSWER: case 1:(9 * 8 * 7 * 6 * 5 * 1) for integers ending in 0. case 2: (8 * 8 * 7 * 6 * 5 * 4) for integers ending in 2,4,6, and 8.
result: case 1 + case 2 = 68,800
- My Questions:
Why do we start with 8 and not 9? Could you please explain how case 2 is structured?