4 digit number be arranged from 7,2,6,8,5,1 and divisible by 5

Question

How many ways can the digits $1,2,5,6,7,8$ be arranged into $4$ digits numbers divisible by $5$ provided in the number digits should not repeat? I don't know how to solve these type of questions but I was getting an answer of $60$. Is this correct?

Answer 1

We have four digits to fill. In order for the final number to be divisible by 5, we have to have 5 as the last digit. All other possible numbers as final digit will result in a number not divisible by 5.

Hence, we can only fill the first three digits with the other numbers. Since we do not want repetitions, we exclude 5 and can put any of 7,2,6,8,1 into the first three digits. Again we do not want repetitions. So we can pick any of the five numbers 7,2,6,8,1 to be the first digit. For the second digit we can only pick a number we haven't chosen before to avoid repetition. This leaves 4 numbers to pick form for the second and three numbers for the third digit.

In total we, thus, have $5\cdot 4 \cdot 3 =60$ possibilities.

Answer 2

So the formula for this is the Binomial Coefficient function, written as $$^nC_k$$ where you want to choose $k$ elements from a set of $n$ elements. The formula for this is $$\frac{n!}{k!(n-k)!}$$ where $!$ represents the factorial function. This assumes that order doesn't matter, however in your case order is important (1265 and 6215 are different answers) so you multiply the answer by the number of ways to arrange $k$ items, which is $k!$. So the final answer is $$\frac{n!(k!)}{k!(n-k)!} = \frac{n!}{(n-k)!} = \frac{5!}{2!} = \frac{120}{2} = 60.$$

Answer 3

There are actually 2 ways that I can think of, but 1 would be more efficient than the other.

1) as pointed out by @YukiJ, we can exclude 5 since 5 needs to be the last number. Hence, there are 5 possibilities for the first number. Then 4 for the second and 3 for the third, which is just 5.4.3 = 60. This works for simple questions like these, but I would prefer the second approach.

2) In the same manner, we have 5 numbers that we can pick from for the first 3 digits, and we want to be able to see all the combinations that we can swap them around. Hence, this will give 5C3 (i.e choosing any 3 numbers out of 5), and then multiply it by 3! (=6), which means all the ways you can jumble up the 3 numbers that you pick. Take note that this 3! works because all the numbers that you have are unique, otherwise you will need to divide it by other numbers, which I will not go into.