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I am trying to find out the number of 4 digit numbers formed from digits 0 to 9 without repeating any digit which are divisible by 5? But dont know how to do it as if last digit ends with digit 0 then there will be 9 possible digits at first place(excluding 0).. But if last digit ends with digit 5 then there will be 8 possible digits at first place(excluding 0 and 5) So how to do it in easiest way?

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First case when the last digit is 0

now you don't want repetition thus possible numbers of 4 digit numbers divisible by 5 is $9*8*7 $

Now the second case when the last digit is 5 but we also want the first digit should be non-zero

Thus filling the first place we have only 8 choices because we exclude zero

for second place we also have 8 choices

now for third place, we have 7 choices

Thus for second case total $=8*8*7$

$$\text{Total}=9\cdot8\cdot7+8\cdot8\cdot7=56\cdot17=952$$

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  • $\begingroup$ So that will be 952 right? $\endgroup$ – manish thakur Oct 2 '18 at 14:25
  • $\begingroup$ Yes The answer is 952 but did you understand what I am trying to say? $\endgroup$ – Deepesh Meena Oct 2 '18 at 14:28
  • $\begingroup$ +1 for clear explanation. This is exactly same as my attempt. $\endgroup$ – Avinash N Oct 2 '18 at 14:28
  • $\begingroup$ Yeah i got it deepesh. I was trying to solve it without taking different cases. But i got it now. $\endgroup$ – manish thakur Oct 4 '18 at 13:43
  • $\begingroup$ is it possible that you can accept it as answer and clear the queue $\endgroup$ – Deepesh Meena Oct 6 '18 at 20:01
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Work backwards... There are two cases, either the units digit is $0$ or $5$. If it ends in $0$, then the tens digit can be $1-9$, 9 possibilities, the hundreds digit can be the remaining $8$, and the thousands can be the remaining $7$. So the total is $9\cdot 8\cdot 7$.

Note then if you start with $5$ in the units place, you approach the same way, however, the thousands digit can not be $0$ (why?). So the total then is $8\cdot 8\cdot 7$.

Now what do you do with both of these numbers?

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  • $\begingroup$ +1 for detailed explanation. $\endgroup$ – Avinash N Oct 2 '18 at 14:31

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