Arrangements: Forming Numbers Without Repetition

The first decision in every counting question is does order matter? Forming a number, the digits sit in positions, so order matters, and this is a permutation. The slot-by-slot method makes the count transparent.

How many different $4$-digit numbers can be formed from the digits $1, 2, 3, 4, 5, 6, 7$ if no digit may be repeated? [3]

The working

Method 1, fill the four positions in turn. Each choice uses up a digit, so the pool shrinks: $\underbrace{7}_{\text{1st}} \times \underbrace{6}_{\text{2nd}} \times \underbrace{5}_{\text{3rd}} \times \underbrace{4}_{\text{4th}} = 840 \quad \text{(M1, M1, A1)}$

Method 2, the permutation formula (order matters, choosing $4$ from $7$): $^{7}P_{4} = \frac{7!}{(7-4)!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840$

Both give $\mathbf{840}$.

Where the marks are won and lost

• Recognise that order matters (a number’s digits are positional), so this is $^{7}P_{4}$, not $^{7}C_{4}$. Using combinations would undercount by a factor of $4! = 24$.
• The shrinking pool ($7, 6, 5, 4$) reflects “no repetition”. If repetition were allowed it would be $7^4$ instead.
• Stop at four factors, one per digit position. A common slip is running all the way to $7 \times 6 \times \dots \times 1$.

Common mistakes

• Using $^{7}C_{4} = 35$ (treating it as an unordered selection).
• Allowing repetition and computing $7^4 = 2401$.
• Multiplying the wrong number of factors.

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