Revise Counting principles, permutations and combinations for IB Mathematics: Analysis and Approaches (1.4)

Almost every counting problem is built from two ideas. The product (multiplication) principle handles a task done as a sequence of stages carried out together: if there are $m$ ways to complete the first stage and, for each of these, $n$ ways to complete the second, then there are $m \times n$ ways to complete both. The addition principle handles a choice between separate, non-overlapping cases: if an outcome can arise in $m$ ways in one case or $n$ ways in a second case, with no overlap, there are $m + n$ ways in total. The quick test is the connecting word: 'and' between stages means multiply; 'or' between mutually exclusive cases means add.

A permutation is an arrangement of objects in a definite order. Arranging books on a shelf, assigning medals, setting a password — in each case changing the order produces a genuinely different outcome. To count the arrangements of $n$ distinct objects we use the factorial $n!$, the product of every positive integer up to $n$. There are $n$ choices for the first position, then $n-1$ for the second, and so on down to $1$, which is exactly $n!$. When we arrange only $r$ of the $n$ objects, the count is $^nP_r$.

$n! = n \times (n-1) \times (n-2) \times \cdots \times 1, \qquad 0! = 1$

^nP_r = \dfrac{n!}{(n-r)!}

Many arrangement questions add a condition: a particular item must go in a fixed place, two items must stay together, or an item is barred from a position. The reliable strategy is to deal with the restriction FIRST and count the rest afterwards. If a specific item must occupy a position, place it, then arrange what remains. If two items must be adjacent, glue them into a single block, arrange the blocks, then multiply by the internal arrangements of the block. If an item is forbidden somewhere, either count the allowed placements first or subtract the forbidden cases from the unrestricted total.

A combination is a selection where order is irrelevant. Forming a committee, dealing a hand of cards, choosing lottery numbers — the group you end up with is all that matters, not the sequence in which you picked it. Because each unordered group of $r$ items could be arranged in $r!$ different orders, there are always $r!$ times fewer combinations than permutations, which is exactly why we divide by $r!$ in the formula.

$^nC_r = \dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!}, \qquad ^nP_r = {}^nC_r \times r$!

When a question asks for 'at least one' of something, counting each qualifying case separately — exactly one, exactly two, and so on — is slow and a frequent source of error. The efficient method is the complement: count the total number of selections, then subtract the single case you do NOT want, which is usually 'none'. What remains is everything with at least one. This turns a multi-case sum into a single subtraction.

So far every object has been distinct. When some items are identical, swapping two of them does not create a new arrangement, so the plain factorial over-counts. To correct this, divide $n!$ by a factorial for each group of identical items. Arranging $n$ objects, where one type appears $p$ times and another appears $q$ times, gives $\dfrac{n!}{p!\,q!}$; the division removes the arrangements that differ only by rearranging indistinguishable items.

$\text{Distinct arrangements} = \dfrac{n!}{p!\,q!\,\cdots} \quad\text{where } p, q, \ldots \text{ count each repeated type}$

In an IB Mathematics question the marks are analytic: each is tied to a specific line of working. A method mark (M) rewards a correct approach — here, recognising that order does not matter and setting up the right product of combinations — and an answer mark (A) rewards the correct value, but an A-mark is dependent on the M-mark that earns it. Follow-through (FT) means a value that follows correctly from your own earlier (possibly wrong) figure still scores, in-between working is ignored (ISW) once a correct answer is seen, and equivalent forms are accepted. Study how each of the 4 marks below is pinned to one line.

Counting is the engine room of probability. Once you can count the total outcomes and the favourable outcomes, a probability is simply the ratio of the two, and combinations reappear throughout the binomial distribution, where $\binom{n}{r}$ counts the ways of getting $r$ successes in $n$ trials. The binomial theorem in algebra uses the very same $\binom{n}{r}$ as its coefficients. Master the core habit here — decide whether order matters, decide whether to multiply or add, deal with restrictions first, and show every line — and these later topics become variations on a method you already own.