Combinations and Permutations | Free Exam P Lesson

Pick 6 from 49 in a lottery. Count arrangements instead of subsets and you get a number 720 times too large, because $P (6, 6) = 6! = 720$ orderings map to the same ticket.

If a procedure has $k$ sequential stages with $n_{1}, n_{2}, \dots, n_{k}$ choices respectively, the total number of outcomes is:

$n_{1} \times n_{2} \times \dots \times n_{k}$

This applies whenever choices at each stage are independent of earlier choices.

A permutation is an ordered arrangement. Choose $r$ objects from $n$ distinct objects:

$P (n, r) = \frac{n !}{( n - r )!}$

Special case: arrange all $n$ objects, that gives $n!$ .

When $n$ objects include groups of identical items ( $n_{1}$ of type 1, $n_{2}$ of type 2,..., $n_{k}$ of type $k$ ) the distinct arrangements form the multinomial coefficient:

$\frac{n !}{n _{1} ! \cdot n _{2} ! \dots n _{k} !}$

MISSISSIPPI has 11 letters: 1 M, 4 I's, 4 S's, 2 P's. Distinct arrangements: $11! / (1! \cdot 4! \cdot 4! \cdot 2!) = 34, 650$ .

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Common mistakes

Using permutations when combinations are needed. Choosing 3 engineers from 7 with $P (7, 3) = 210$ overcounts by $3! = 6$ . Trap: 210 instead of 35.
Forgetting to divide by repeated elements. Arrangements of MISSISSIPPI as $11! = 39, 916, 800$ instead of $34, 650$ . Trap: 39,916,800.
Missing cases in "at least" problems. Computing only "exactly 1 green" and forgetting "exactly 2 green." Trap: 120 instead of 150.

Bottom line

Fundamental counting principle: multiply the number of choices at each stage; it underlies every permutation and combination formula.
Permutations (order matters): $P (n, r) = n! / (n - r)!$ . Combinations (order doesn't): $(r n) = n! / (r! (n - r)!)$ ; they differ by a factor of $r!$ .
Key question: does rearranging the selected objects create a different outcome? Yes is a permutation, no is a combination.
Multinomial coefficient: $n! / (n_{1}! \cdot n_{2}! \dots n_{k}!)$ arranges objects with repeated identical copies (MISSISSIPPI gives 34,650).

Exam shortcut

When stuck, translate the problem into a physical process: "I am pulling 4 balls, do I care about the order I pull them?" This forces the permutation vs. combination decision before you write a formula. "Combinations are permutations with the ordering divided out." $(r n) = P (n, r) / r!$ . If you accidentally compute a permutation, dividing by $r!$ recovers the combination.

The full lesson (about 2,618 words, 17 min read) adds 3 worked examples, all 7 common mistakes, a self-check, free in the app.

Learning objectives

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