Permutations of 3 digit numbers divisible by 5

Question

I recently had to answer the following permutation question:

How many 3 digit numbers can be formed from the digits 2,3,5,6,7,9 which are divisible by 5 and none of the digits are repeated?

Having not done this kind of algebra since high school could someone break down how this would be calculated. I'm assuming to calculate the number of 3 digit combinations it's half the permutations = 6 x 5 x 4?

Where does the $5$ have to be for the number to be a multiple of $5$? Does it matter where any of the other numbers go after that? — abiessu, Apr 25 '16 at 12:31
Technically these are not permutations but sequences of $6$ objects of length $3$ . For the answer, I quote @abiessu , think of the property of being divisible by $5$. — Oscar, Apr 25 '16 at 12:35

score 4 · Accepted Answer · answered Apr 25 '16 at 12:44

4

Choose $2$ digits from $[2,3,6,7,9]$ and arrange them in any order: $\binom52\cdot2!=20$.

answered Apr 25 '16 at 12:44

barak manos

43,109

Can you add more of an explanation. Why 2 numbers? – Ben Foster Apr 25 '16 at 13:12
1

@BenFoster: The third digit has to be $5$ in order for the number to be divisible by $5$. So you can only choose the first digit and the second digit. – barak manos Apr 25 '16 at 13:13

Permutations of 3 digit numbers divisible by 5

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