Forming Numbers with a Restriction

The rule for any counting problem with a restriction is: fill the restricted position first. An “even” number is restricted at the last digit (it must be even), so you count that position’s choices before the free ones. This “restricted position first” approach prevents overcounting and handles most 0606 restriction questions.

How many different $4$-digit even numbers can be formed from the digits $1, 2, 3, 4, 5, 6$ if no digit is repeated? [4]

The working

Step 1, deal with the restriction, the last digit must be even. The even digits available are $2, 4, 6$, so there are $3$ choices for the last position: $\text{last digit: } 3 \text{ choices} \quad \text{(M1)}$

Step 2, fill the remaining three positions from the $5$ digits left (one even digit is now used), with no repetition: $\underbrace{5}_{\text{1st}} \times \underbrace{4}_{\text{2nd}} \times \underbrace{3}_{\text{3rd}} = 60 \quad \text{(M1)}$

Step 3, multiply (last digit choices $\times$ arrangements of the rest): $3 \times 60 = 180 \quad \text{(M1, A1)}$

Where the marks are won and lost

• Restricted position first. Choosing the last (even) digit before the others is what keeps the count correct. Filling left-to-right and imposing “even” at the end causes overcounting or double-handling.
• After using one even digit for the last place, only 5 digits remain for the first position, not 6. The pool shrinks because there’s no repetition.
• The three free positions are a permutation: $5 \times 4 \times 3$, order matters (it’s a number).

Common mistakes

• Filling the first digit first and struggling to impose “even” afterward.
• Using $6 \times 5 \times 4$ for the free positions (forgetting one even digit is already used).
• Adding instead of multiplying the restricted and free counts.

Full method: The Counting Principle notes. See also Permutations (nPr). Topic home: Permutations & Combinations pillar.