How many four digit numbers can formed using digits 0 to 7 which is divisible by 4 without repetition?
We can use digits from 0 to 7only. Following are the possibilities.
04, 12, 16, 20, 24, 32, 36, 40, 52, 56, 60, 62, 64 – total 13
A number can’t start with zero, therefore cases above having a 0, will have to be dealt separately.
04,20,40,60 – 4possibilities
For first 2 digits we can use product rule to get 6 × 5=30. Because, out of 8digits 2 have been already used up. So, in all, there are 30×4=120possibilities.
For the remaining 9 cases, the first 2digits can be chosen in 5×5=25 Because, out of remaining 6digits, we can’t choose 0 as the starting digit. So, there are
25 ×9=225
So, combining the above cases, we get the answer to be 120+225=345
i dont understand what i am unable to get here. help me to sort out this.
Best Answer
There are 4 even number and 4 odd. 2 are divisible by 4 (0 and 4) and 2 are even not divisible by 4 (2 and 6).
100 is divisible by4 so $xxAB$ is divisible by 4 iff $AB$ is divisible by 4.
$AB$ is divisible by $4$ if i)$B$ is divisible by 4 and $A$ is even. $2*(4-1) = 6$ possibilities for this. or ii)$B$ is even but not divisible by 4 and $A$ is odd. $2*4 = 8$ possibilities for this.
So there are $14$ possibilities of $AB$. $cdAB$ may not have $c = 0$. If $AB$ are both not $0$ then there are $5$ choices for $c$ and $5$ for $d$ so $25$ choices. If $A$ or $B$ is $0$ there are six choices for $c$ and $5$ for $d$ so $30$ choices.
If $B = 0$ there are $3$ even numbers that $A$ can be. If $A=0$ then $B$ is one number divisible by $4$ that $B$ can be. So there are $4$ choices for $AB$ with a zero and $10$ choices where neither are zero.
So there are $4*30+10*25 = 370$ total choices.
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The only thing you did wrong was i) put 62 in your list ii) leave 72 and 76 out of the list.