Find the number of three-digit numbers in which all the digits are distinct, odd and number is a multiple of 5.

Question

So, I tried to solve like this:

One's place: 5 (since the number is a multiple of 5 and odd); Ten's place: the rest 4 odd/even numbers + a zero; Hundred's place: the remaining 4 odd/even numbers

So the required number of 3 digit natural odd numbers, which are also a multiple of 5, should be: 4 * 5 * 1 = 20

But the book I use says it should be 12.

and a quick google search also told me it should be 12. (which I am not convinced is the right answer)

Like this:

number of ways of filling unit place is only one i.e. 5. Now, four odd digits are left, hence ten’s place can be filled in four ways and hundred’s place in three ways. number of required three-digit natural numbers is 1 x 4 x 3 = 12

So now, I understand up to the part that it says filling up the ten's place with the four remaining odd digits. After that it says that "hundred's place can be filled in three ways." But there are five digits remaining? (4 even numbers + a zero)

Now, to my question. Did I get the right answer(20)? Or am I missing something here?

Also, I have seen the other related posts here but they seem to lead into a completely different answer, like 64 [8(hundred's) * 8(ten's) * 1(one's)]

Thank you, in advance.

Answer 1

You can think of it as a permutation problem:

A three digit number that is a multiple of 5 and is odd must end with 5, since it can't end in 0. So you need all possible 2-permutations for digits (1,3,7,9). In this way you get the first 2 digits of the number we are intested in, sticking the 5 in the last position, as you said.These are 4 *3 possible permutation, In this way you are sure you use only odd digits for the permutations, as the ex. requires