Combinations (n Choose r)

Statement

A combination counts how many different ways we can choose \(r\) items from \(n\) distinct items when order does not matter. It is read “n choose r” and written \(\binom{n}{r}\). The formula is:

\[\binom{n}{r}=\frac{n!}{r!\,(n-r)!}\]

Here, \(n!\) (pronounced “n factorial”) means the product \(n\times(n-1)\times\cdots\times 2\times 1\), with \(0! = 1\) by definition. The value \(\binom{n}{r}\) is always a whole number because it counts selections.

Why it’s true (short reason)

• If order did matter, there would be \(nP r = \frac{n!}{(n-r)!}\) ways (arrangements/permutations).
• But each unique selection of \(r\) items can be ordered in \(r!\) ways that are considered the same when order doesn’t matter. So we divide by \(r!\).
• Therefore \(\binom{n}{r}=\dfrac{n!}{(n-r)!}\times\dfrac{1}{r!}=\dfrac{n!}{r!(n-r)!}\).

Recipe (how to use it)

1. Confirm that order does not matter (e.g., a committee, a hand of cards, choosing toppings where the list order is irrelevant).
2. Identify \(n\) (total distinct items) and \(r\) (how many are chosen).
3. Compute with \(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\). Cancel factors before multiplying to keep numbers small.
4. Use the symmetry \(\binom{n}{r}=\binom{n}{n-r}\) to pick the smaller of \(r\) and \(n-r\) if it helps.

Spotting it

Typical signals: “choose”, “select”, “committee of”, “hand of”, “no order”, “combinations”. If the wording mentions “arrangements”, “orders”, or “line-ups”, you probably need permutations instead.

Common pairings

• Permutations: \(nP r=\dfrac{n!}{(n-r)!}\) when order matters (e.g., podium places).
• Binomial probabilities: The number \(\binom{n}{r}\) appears in \(P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}\).
• Counting with constraints: Choose groups in stages and multiply (rule of product), often using combinations in each stage.

Mini examples

1. Choose 3 from 7: \(\binom{7}{3}=\dfrac{7\cdot6\cdot5}{3\cdot2\cdot1}=35\).
2. Choose 1 from 20: \(\binom{20}{1}=20\) (you just pick which one).
3. Choose 0 from n: \(\binom{n}{0}=1\) (there is exactly one empty selection).

Worked reasoning tools

• Cancellation first: Write factorials as products and cancel numerators/denominators before multiplying. Example: \(\binom{12}{2}=\dfrac{12\cdot11}{2}=66\).
• Symmetry: \(\binom{n}{r}=\binom{n}{n-r}\). Use whichever has the smaller second number. Example: \(\binom{20}{18}=\binom{20}{2}=190\).
• Multiplying stages: If a selection is split into categories, multiply the counts for each independent stage. Example: choose 2 girls from 5 and 3 boys from 7: \(\binom{5}{2}\binom{7}{3}\).
• At least/at most: Sum the relevant values: “at most 2” means \(r=0,1,2\). “At least 1” is total minus the \(r=0\) case.

Pitfalls

• Using permutations by mistake: If order is irrelevant, do not use \(nP r\). Switch to \(\binom{n}{r}\).
• Large factorials un-simplified: Don’t compute full \(n!\). Cancel common factors early to avoid huge numbers.
• Wrong \(n\) or \(r\): Carefully identify the total pool and how many are actually chosen.
• Forgetting symmetry: Calculating \(\binom{n}{n-1}\) directly is slow; it equals \(n\).

Exam strategy

• Underline “order does not matter”. If you see “order matters”, use permutations instead.
• Write the combination formula and plug in numbers carefully.
• Cancel before you multiply; use symmetry to keep arithmetic light.
• For multi-part selections, break into stages and multiply, checking that stages are independent and cover the constraint.

Summary

Combinations count selections where order is irrelevant. The formula \(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\) derives from taking all \(r\)-permutations and dividing by the \(r!\) internal arrangements that do not create different selections. In practice, you keep numbers manageable by cancelling factors, using symmetry to pick the smaller of \(r\) and \(n-r\), and multiplying independent stages for mixed-category problems. This tool underpins many GCSE counting and probability questions, especially binomial ideas and structured selection problems.