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

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)]

• Since 5 has to be the last digit, there are 4 digits left 1,3,7,9. All the combinations from these numbers are $\binom{4}{2}$=6. Every combination has two possible permutations, so the total number we want is 2x6=12
– user1054388
Commented Jun 30, 2022 at 17:12

• It is just (excluding five, last digit) $\,2\binom{4}{2}$=12