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Determine how many $4$ digit numbers divisible by $5$ can be generated using the set: $\{1,3,5,6,7,8\}$. Repetion is not allowed.

If I am not wrong, we have to use permutation in this question right? I still do not understand how to solve this question. I know that the numbers must end with either $5$ or $0$. Looking at the question I can also tell that there are $5$ possibilities ($1,3,6,7$, or $8$) for the thousands digit, $4$ possibilities for the hundreds digit, $3$ possibilities for the tens digit, and $1$ possibility for the ones digit. I just don't know how to use the formula, and how to show my work. I need someone to show me all the steps, and the answer, so I can use this to solve similar questions. Can someone please help me?

vitamin d
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Muhammad Khan
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    Welcome to Mathematics Stack Exchange. Note that a number divisible by $5$ ends with $5$ or $0$ – J. W. Tanner Jul 25 '21 at 20:13
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    A number is divisible by 5 if and only if the last digit is 5 or 0. You are correct, that you will need permutation. So your numbers look like this: xxxxx5. Now how many ways are there to place the first 5 digits? – Cornman Jul 25 '21 at 20:20
  • Hi Muhammad. Can you tell us, what is your idea of numbers generating other numbers? (i.e., how does it work?) You haven't mentioned any operation, besides division. – 311411 Jul 25 '21 at 20:21
  • @Cornman: per the title, OP is looking for $4$-digit numbers – J. W. Tanner Jul 25 '21 at 20:23
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    @J.W.Tanner Thanks for pointing that out. I overlooked that. The method still remains to be kinda the same. Also I think the hint that 5 has to be the last digit is what Muhammed Khan missed. – Cornman Jul 25 '21 at 20:24
  • @all I do know that the number must end with a 5 or 0. But, can someone show me the steps of using permutation and how I can get the answer. Permutations are really hard for me, and if someone can show me all the steps It'll be very helpful. – Muhammad Khan Jul 25 '21 at 20:26
  • @MuhammadKhan: You should put what you know in the question -- not in comments. There are $5$ possibilities $(1, 3, 6, 7, $ or $8)$ for the thousands digit, $4$ possibilities for the hundreds digit, $3$ possibilities for the tens digit, and $1$ possibility (namely, $5$) for the ones digit. Can you take it from here? – J. W. Tanner Jul 25 '21 at 20:31
  • @J.W.Tanner I edited my question. – Muhammad Khan Jul 25 '21 at 20:49

2 Answers2

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Any multiple of $5$ generated from that set must have a $5$ at the end.

Now, the rest three digits have to be chosen from the set $\{1,3,6,7,8\}$. So, the first digit has $5$ choices, the second one has $4$ choices, and the third one has $3$ choices.

So, the total number of ways is $$5\times 4\times 3=60$$

Does that help?

Sayan Dutta
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  • Yes, that helps a lot. But, don't we have to use the permutation formula? Is this all the work we would show? I am worried because this is a question from my final assignment and there are a few questions similar to this, If I can understand how to solve this the rest should be easy. – Muhammad Khan Jul 25 '21 at 21:11
  • @MuhammadKhan According to me, this is enough for the question you gave. I don't think we need the permutation formula here. You use that formula when you need to permute things, i.e., you want to arrange things. But, here the arrangement is already taken care of. – Sayan Dutta Jul 25 '21 at 21:15
  • Alright. Thank you! – Muhammad Khan Jul 25 '21 at 21:17
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You've almost got the answer to the question.

Just use the multiplication principle.

If you want to use permutations, it's $P(5,3)$.

J. W. Tanner
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