How many even numbers less than $500$ can be formed using the digits $1$, $2$, $3$, $4$ and $5$

combinationspermutations

How many even numbers less than $500$ can be formed using the digits $1$, $2$, $3$, $4$ and $5$?

Each digit may be used only once in any number.

I first tried by using the formula for permutations:

$$
3P1 \times 6 = 18
$$

But the answer is $28$.

So I tried again, this time by listing all of the possible permutations:

  1. $132$
  2. $142$
  3. $152$
  4. $124$
  5. $134$
  6. $154$
  7. $214$
  8. $234$
  9. $254$
  10. $312$
  11. $342$
  12. $352$
  13. $314$
  14. $324$
  15. $354$
  16. $412$
  17. $432$
  18. $452$

I still get $18$.

What am I doing wrong?

Best Answer

Case $1$: The number is a single digit.

In this case, the only even numbers are $2$ and $4$, giving a total of $2$.

Case $2$: The number has exactly two digits.

In this case, the last digit must be either $2$ or $4$, and the first digit must be one of the other four digits allowed, giving a total of $2 \cdot 4=8$.

Case $3$: The number has exactly three digits.

In this case, the last digit must be either $2$ or $4$ and the middle digit must be one of the other four digits allowed.

If the middle digit is $5$, then the first digit must be one of the other three digits allowed, giving a total of $2 \cdot 3=6$.

If the middle digit is not $5$, then the first digit must be one of two digits other than the last two digits and $5$, giving a total of $2 \cdot 3 \cdot 2=12$.

So, there are $2+8+6+12=28$ even numbers less than $500$ with distinct digits $\in \{1,2,3,4,5\}$.

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