The digits \(1\), \(2\), \(3\), \(4\), \(5\) are written down at random to form a five-digit number.

Note; no digit may be repeated!


  1. how many such numbers are possible,

We have five ways to choose the first digit, and once we’ve done that, four ways of choosing the second, and so on…

So there are \(5! = 120\) possible numbers.

  1. the chance that the last digit is odd,

The last digit must be one of \(1, 3\) and \(5\). Once we have chosen this, there are \(4!\) ways to choose the first four digits.

Thus there are \(3 \times 4! = 72\) possible numbers that are odd and the chance that a number is odd is \(\dfrac{72}{120}=\dfrac{3}{5}.\)

Could we have just stated this anyway as \(3\) of our digits were odd?

  1. how many of the numbers are divisible by \(4\).

Any five-digit positive integer can be written as \(100a + b\) for positive integers \(100 \le a < 1000\) and \(0 \le b < 100\).

Then we only need to check \(b\) for divisibility by \(4\) because \(100a + b = 4\left(25a + \frac{b}{4}\right)\) which is an integer if \(b\) is a multiple of \(4\).

So the general rule for checking whether or not \(4\) goes into a number is this; does \(4\) go into the final two digits?

If so, then the number is a multiple of \(4\), and not otherwise. (That’s how we check for leap years and the next Summer Olympics.)

The only pairs of digits that could end our number and which are divisible by \(4\) are \(12\), \(24\), \(32\) and \(52\).

Thus, there are \(3! \times 4 = 24\) such numbers.