Mathematics · Ch 5 — Permutations and Combinations

Derivation of the Formula for nPr

5.3.3

Derivation of the Formula for nPr

From the Product Form to the Factorial Form

The earlier formula for the number of permutations of $n$ different objects taken $r$ at a time was a product of $r$ decreasing factors:

^{n}P_{r} = n(n-1)(n-2)\cdots (n-r+1)

This form is correct but not compact. To turn it into a neat factorial expression, multiply and divide by the product of all the integers from $(n-r)$ down to $1$ .

Note

Why multiply by that particular string?

The product $(n-r)(n-r-1)\cdots 3\cdot 2\cdot 1$ is exactly $(n-r)!$ . Multiplying numerator and denominator by it does not change the value — it simply fills in the missing factors so that the numerator becomes a complete factorial.

Write the numerator as:

n(n-1)(n-2)\cdots (n-r+1) \times (n-r)(n-r-1)\cdots 3\cdot 2\cdot 1

The whole numerator is now the product of all integers from $n$ down to $1$ , which is $n!$ . The denominator is $(n-r)!$ . Therefore:

^{n}P_{r} = \frac{n!}{(n-r)!}

This holds for $0 < r \le n$ .

The Special Case $r = n$

When $r = n$ , the formula gives:

^{n}P_{n} = \frac{n!}{(n-n)!} = \frac{n!}{0!}

Since $0! = 1$ by definition, we get $^{n}P_{n} = n!$ , which matches the number of ways to arrange all $n$ objects.

Extending the Formula to $r = 0$

What about arranging no objects at all? There is exactly one way to do nothing — leave all objects untouched. So $^{n}P_{0}$ should equal $1$ .

Check the factorial formula for $r = 0$ :

^{n}P_{0} = \frac{n!}{(n-0)!} = \frac{n!}{n!} = 1

The formula works perfectly for $r = 0$ as well. Hence the complete, universally valid statement is:

^{n}P_{r} = \frac{n!}{(n-r)!}, \quad 0 \le r \le n

When Repetition Is Allowed …

Theorem 2

Theorem 2 (Permutations with Repetition Allowed)

n^{r}

The number of permutations of $n$ different objects taken $r$ at a time, where repetition of any object is allowed, is $n^{r}$ .

Hypotheses:

The $n$ objects are all distinct from one another.
We are selecting $r$ objects in order (a permutation, not a combination).
After each selection, the chosen object is returned to the pool, so it can be chosen again in the next position.
$r$ can be any non‑negative integer; there is no restriction $r \leq n$ because repetition makes it possible to use the same object more than once.

When is this used?

Whenever you are counting ordered arrangements (like passwords, PIN codes, or words) where the same item can appear multiple times — for example, the number of 3‑letter words you can form from the letters of the word NUMBER if repetition is allowed is $6^{3} = 216$ .

›Proof

Proof.

We have $n$ different objects, and we need to fill $r$ positions in a sequence.

Step 1 – First position.

There are $n$ choices for the object that goes in the first place.

Step 2 – Second position.

Because repetition is allowed, the object we used in the first position is still available. So for the second position we again have $n$ choices.

Step 3 – Third position.

The same reasoning applies: the object used in the first two positions is not removed from the pool, so we still have $n$ choices for the third position.

Continuing in this way for all $r$ positions, each position independently offers $n$ possibilities.

Step 4 – Applying the multiplication principle.

The total number of ways to fill all $r$ positions is the product of the number of choices for each position:

$n \times n \times n \times \cdots \times n \quad (r \text{ factors}) = n^{r}.$

This completes the proof. ∎ …

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Derivation of the Formula for nPr