Irish Leaving Certificate Mathematics

Counting problems ask how many outcomes an experiment has, or how many ways something can be arranged or chosen, without listing every possibility. The same ideas underlie the probability of equally likely outcomes, where the probability of an event is the number of outcomes in the event divided by the total number of outcomes.

Listing outcomes systematically

The outcomes of a simple experiment can be listed in an organised way so that none is missed or repeated. For two coins the outcomes are HH, HT, TH and TT. For two fair dice the outcomes form a $6 \times 6$ table of 36 ordered pairs, where $(2, 5)$ and $(5, 2)$ are different outcomes. Listing in order (all outcomes beginning with the first possibility, then the second, and so on) or drawing a table or tree diagram are the usual ways of keeping a list complete.

The fundamental principle of counting

If one task can be done in $m$ ways and, whichever way it is done, a second task can then be done in $n$ ways, the two tasks together can be done in $m \times n$ ways. The rule extends to any number of tasks done one after another.

A restaurant offers 4 starters, 6 main courses and 3 desserts. A full meal with one of each can be chosen in $4 \times 6 \times 3 = 72$ ways.

The principle applies just as well when the choices at later stages depend on earlier ones, provided the number of options at each stage is fixed. A PIN of 4 digits, where the first digit cannot be 0 and no digit may repeat, can be formed in $9 \times 9 \times 8 \times 7 = 4\,536$ ways: the first digit has 9 options (1 to 9), the second has 9 (any digit except the one used, and 0 is now allowed), the third 8 and the fourth 7.

Factorials and arrangements of distinct objects

The number of ways of arranging $n$ distinct objects in a row is found by choosing the first position ($n$ options), then the second ($n - 1$), and so on down to 1:

read "$n$ factorial". By convention $0! = 1$, which keeps the formulas below working when nothing is left to arrange. Since $n! = n \times (n-1)!$, a ratio of factorials can often be simplified by cancelling the common part:

Five different books can be placed on a shelf in $5! = 120$ ways.

Arrangements with conditions

Conditions on positions are handled by placing the restricted objects first.

Seven people stand in a line and two particular people, Aoife and Brian, must stand next to each other. Treat the pair as one block. The block and the other 5 people make 6 units, arranged in $6! = 720$ ways, and within the block Aoife and Brian can swap, giving $720 \times 2 = 1\,440$ arrangements.

If instead Aoife and Brian must not be next to each other, subtract the cases where they are together from all the arrangements: $7! - 1\,440 = 5\,040 - 1\,440 = 3\,600$.

A number of 5 digits is made from the digits 1, 2, 3, 4, 5, each used once, and must be even. The last digit must be 2 or 4, so there are 2 choices for it, then $4!$ ways to arrange the remaining four digits, giving $2 \times 24 = 48$ numbers.

Arranging r objects from n

The number of ways of arranging $r$ objects chosen from $n$ distinct objects, where order matters, is the product of the first $r$ factors of $n!$:

Three of eight runners finish first, second and third, so the podium can be filled in ${}^8P_3 = 8 \times 7 \times 6 = 336$ ways. Note that ${}^nP_n = n!$, and ${}^nP_1 = n$.

Selections: combinations (Higher level only)

Higher level only: when order does not matter, the number of ways of selecting $r$ objects from $n$ distinct objects is written ${}^nC_r$ or $\binom{n}{r}$. Each selection of $r$ objects can be arranged in $r!$ ways, so the number of arrangements is $r!$ times the number of selections, and therefore

These numbers are called binomial coefficients because they are the coefficients in the expansion of $(a + b)^n$. Some useful facts follow from the formula:

The last says that choosing $r$ objects to take is the same as choosing the $n - r$ to leave behind. Also

which is the rule that builds Pascal's triangle, each entry being the sum of the two above it.

A committee of 3 is chosen from 10 people: $\binom{10}{3} = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$ committees. Working with the smaller of $r$ and $n - r$ keeps the arithmetic short.

Combined selections

When a selection has independent parts, multiply the counts for each part. A team of 5 is to have 3 girls from 7 and 2 boys from 6:

When a condition says "at least", it is often easier to count the complement. From 8 people (5 men and 3 women) a group of 4 is chosen with at least one woman. All groups number $\binom{8}{4} = 70$, and the groups with no women number $\binom{5}{4} = 5$, so there are $70 - 5 = 65$ groups.

Choosing and arranging

Selecting and then arranging can be combined. Choose 3 of 6 people, then seat them in a row: $\binom{6}{3} \times 3! = 20 \times 6 = 120$, which equals ${}^6P_3 = 6 \times 5 \times 4 = 120$, as it must.

Counting and probability

When all outcomes are equally likely, a count of favourable outcomes divided by a count of all outcomes gives a probability. A hand of 2 cards is dealt from a pack of 52. The number of possible hands is $\binom{52}{2} = 1\,326$ and the number of hands made of two aces is $\binom{4}{2} = 6$, so