I found a question and I have different approach to solve it , but unable to get the answer.
Question :How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
I solve it like this :
divisible by 5 : means I have to put '5' at unit digit.
now I have 5 numbers left(excluding '5') , and have to choose 2 of them.
I just learn permutation and combination and found that this is the example of permutation.(reason : 21 is not equal to 12).
so `5P2 ways to solve this. which makes my answer to 5*4*3=60.
their solution :
divisible by 5 : means I have to put '5' at unit digit.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
so the answer is going to be 5*4=20
please tell me where I am missing. any topic name or link to the tutorial is welcomed.
So there are 3 digits right? Consider the number be called:
$u = xyz$ where $xyz$ are just digits (Not multiplied)
You know $u$ is divisible by 5 therefore $z = 5$ is certain. No numbers can be repeated.
Total = 6 numbers. Because $z = 5$, total left = 5 numbers.
$u = xy5$
There are 5 remaining possibilities of $y$ There are now 4 remaining possibilities of $x$
There is ONE (1) possibility of $z$ $u = xy5$
Total Possibilities: (5)(4)(1) = 20 possibilities.