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Here is my approach:- Firstly, I fixed the last digit as $4$ then there will be only $2$ numbers $(0,2)$ for the ten's digit, $3$ numbers for the hundred's digit and $2$ numbers for the Thousand's digit (so that they don't repeat). Number of $4$ digit numbers in which $4$ is the last digit and is divisible by $4 = 2 \times 3 \times 2 = 12$. As there can be only $4,2,0$ as the last digit so there are $12\times 3 = 36$ numbers possible but that is an incorrect answer. The correct answer is $30$. Where did I go wrong?

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    $\begingroup$ The last two digits must be divisible by 4. So, the last digit must be $0,4$ and the second last even, or the last digit must be $2$ and the second last $1$ or $3$. $\\$For the first possibility, there are $2\cdot2\cdot3\cdot2=24$ possibilities. For the second, there are $1\cdot2\cdot1\cdot3=6$. $$24+6=30$$ $\endgroup$ – Don Thousand Oct 28 '19 at 5:30
  • $\begingroup$ @ Don thousand How there are 3 possibilities for the ten's digit if we fix the last digit as 4. The tens digit must be 0 or 2 then. $\endgroup$ – Ali Oct 28 '19 at 5:47
  • $\begingroup$ I'm not doing it in that order. If I order it like that, it'd be $2\cdot3\cdot2\cdot2$ and $1\cdot3\cdot2\cdot1$. $\endgroup$ – Don Thousand Oct 28 '19 at 5:52
  • $\begingroup$ Why did you separately calculate the the possibilities of 0,4 and 2. Why not calculate it altogether ? $\endgroup$ – Ali Oct 28 '19 at 6:20
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Case 1: $\underbrace{**}_{\{1,3,4\}}20 \Rightarrow P(3,2)=3!=6$.

Case 2: $**04 \Rightarrow P(3,2)=3!=6$.

Case 3: $**40 \Rightarrow P(3,2)=3!=6$.

Case 4: $**12=\underbrace{**}_{\{3,4\}}12+\underbrace{*}_{\{3,4\}}012 \Rightarrow P(2,2)+C(2,1)=4$.

Case 5: $**32 \Rightarrow P(2,2)+C(2,1)=4$.

Case 6: $**24 \Rightarrow P(2,2)+C(2,1)=4$.

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  • $\begingroup$ Thank you for your solution. But can you tell where did i go wrong in calculating the answer this way ? $\endgroup$ – Ali Oct 28 '19 at 9:30
  • $\begingroup$ You are saying "I fixed the last digit as 4 then there will be only 2 numbers (0,4) for the ten's digit". No, if you fix last digit as $4$, then the ten's digit cannot be $4$ again, the condition says "without repetition". $\endgroup$ – farruhota Oct 28 '19 at 9:34
  • $\begingroup$ 3 ways to fill the last digit (4,2,0). 2 ways to fill the tens digit for every number we fill in the ones place. 3 ways to fill the hundreds digit and 2 ways to fill the thousands digit which makes 3×2×3×2=36 $\endgroup$ – Ali Oct 28 '19 at 9:37
  • $\begingroup$ my bad I meant to say 2 and 0 if we select last digit as 4. $\endgroup$ – Ali Oct 28 '19 at 9:40
  • $\begingroup$ remember, the thousand's digit cannot be $0$. $\endgroup$ – farruhota Oct 28 '19 at 9:41

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