- 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.

- Question:

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?

  • $\begingroup$ Also it would be great if someone can explain how does this down vote process works ? Like what did I post wrong that people think it should be down voted ? $\endgroup$ – Ciruss Feb 15 '17 at 3:15
  • $\begingroup$ Ciruss, this is the wrong site. This site is for professional mathematicians and their PhD students to discuss questions arising in their research. May I inquire why you post here and not somewhere else (did you look into what MO is about before you posted)? $\endgroup$ – user43208 Feb 15 '17 at 3:17
  • $\begingroup$ Sorry, I am a new user. I am gradually getting used to the website. $\endgroup$ – Lulu Feb 15 '17 at 3:21
  • $\begingroup$ @user43208 I'm also a new comer and searching for exact same question and found this one helpful for me. What is the MO you mentioned? Where else should those question being post? At least this question helped me more than the question you left un-clarified. $\endgroup$ – Mengo Jul 12 '20 at 23:41
  • $\begingroup$ @Mengo I'm glad this question helped you. There seems to be no notice of this anymore, but this question was originally posted at MathOverflow, or MO for short. MathOverflow is the StackExchange site for professional mathematicians, and what happened was that the question was deemed off-topic for the clientele there, and so it was migrated here where it found its proper and happy home. That's what my comment that appears here was about (the comment was originally posted at MO but remained here even after the migration). Make sense? $\endgroup$ – user43208 Jul 13 '20 at 0:31

Call the six digits $D_1, D_2, \ldots, D_6$. In case two, the order in which we are choosing them is $D_6, D_1, D_2, \ldots, D_5$.

  • For $D_6$, there are four choices: 2, 4, 6, 8.
  • For $D_1$, there are eight choices: all nine of 1-9 (can't start with zero!), except $D_6$.
  • For $D_2$, there are eight choices: all ten of 0-9, except $D_6$ and $D_1$.
  • For $D_3$, there are seven choices: all ten of 0-9, except $D_6$ and $D_1$ and $D_2$.
  • For $D_4$, there are six choices: all ten of 0-9, except $D_6,D_1,D_2,D_3$.
  • For $D_5$, there are five choices: all ten of 0-9, except $D_6,D_1,D_2,D_3,D_4$.
  • $\begingroup$ Fantastic my friend, thank you! $\endgroup$ – Lulu Feb 15 '17 at 4:07
  • $\begingroup$ The key to understanding this problem was placing the specific numbers in sets for example D1 , ... , D6 $\endgroup$ – Lulu Feb 15 '17 at 4:09

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