How to count combinations

Many Add Math questions ask 'in how many ways can we choose...' a group where the order of selection makes no difference.

Picking 3 students to form a committee, choosing 2 books from a shelf, or selecting 4 questions to answer are all combinations: the group $\{A,B,C\}$ is exactly the same group as $\{C,B,A\}$. Counting these by writing out every possibility quickly becomes impossible, so we use the combination formula ${}^{n}C_{r}$, read as 'n choose r'.

It tells us precisely how many different unordered selections of $r$ items can be made from $n$ distinct items. This same idea underpins the binomial theorem and the binomial distribution later in Form 5, so a firm grip on counting combinations pays off well beyond this single topic and makes the whole chapter feel far more predictable.

Reach for a combination whenever a question is about choosing or selecting a group and the order inside that group does not matter. Signal words include 'select', 'choose', 'form a committee', 'a team of', 'a group of', or 'how many ways to pick'.

Contrast this with a permutation, where arranging, ordering, ranking, or seating in a row does matter, there the count is ${}^{n}P_{r}$.

A quick test: if swapping two chosen items produces a genuinely different outcome, it is a permutation; if the swap gives back the same selection, it is a combination. Watch too for extra conditions such as 'a particular person must be included' or 'must not be included', which change the numbers you feed into the formula rather than the method itself.

1. 1

   Confirm order does not matter

   Decide whether the question is about selecting a group (combination) or arranging items in order (permutation).

2. 2

   Identify n and r

   $n$ is the number of items available to choose from; $r$ is how many are chosen.

3. 3

   Write the formula

   State ${}^{n}C_{r}=\frac{n!}{r!(n-r)!}$ before substituting.

4. 4

   Substitute the values

   Put your values of $n$ and $r$ into the formula.

5. 5

   Cancel and simplify

   Cancel the larger factorial first, then multiply what remains, do not expand every factorial in full.

6. 6

   Deal with any condition

   If certain items must be included, fix them first and choose the rest from what is left; if some must be excluded, remove them from $n$.

7. 7

   State the whole-number answer

   A count of selections is always a positive whole number.

• Using ${}^{n}P_{r}$ (order matters) when the question is really a combination, this over-counts every group by a factor of $r!$.
• Forgetting the symmetry ${}^{n}C_{r}={}^{n}C_{n-r}$, which can make the arithmetic much shorter.
• With a 'must be included' condition, reducing only $n$ but forgetting to reduce $r$ as well.
• Expanding full factorials on the calculator and slipping, instead of cancelling the larger factorial first.
• Adding results when independent choices should be multiplied, an 'and' between two selections usually means multiply.

The single step that trips students up is deciding whether order matters, many reach for ${}^{n}P_{r}$ out of habit and quietly over-count. In a one-to-one lesson our teachers give you a short, mixed set of 'choose' and 'arrange' questions and watch which route you take, correcting the decision the very moment it goes wrong so the reasoning becomes automatic.

We also show you how to cancel factorials cleanly, which keeps the working tidy and easy to credit line by line. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

To see how we teach combinations, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.