The multiplication principle

Suppose we pick one of $2$ shirts, white or blue, and one of $3$ pairs of trousers: jeans, chinos or shorts. The white shirt can go with any of the $3$ pairs, which gives $3$ outfits. The blue shirt also gives $3$. So there are $3 + 3 = 2 \times 3 = 6$ outfits.

A tree diagram draws a choice like this. It starts at a single point and splits into one branch for each shirt, and each of those branches splits into one branch for each pair of trousers. The endpoints on the bottom row are the leaves, and each leaf is one outcome: the leaf reached through white and then jeans is the outfit $(\text{white}, \text{jeans})$.

Taking the trousers first instead gives $3 \times 2$, the same $6$ outfits. The order in which we take the stages does not change the count.

The same reasoning applies to a choice made in any number of stages.

An ice cream shop offers cake cones and waffle cones, and chocolate, vanilla and strawberry ice cream. One cone is ordered with one flavour. Count the possible cones, taking the cone first and then taking the flavour first.

Solution

Taking the cone first, the choice has two stages:

• Stage $1$, the cone: cake or waffle, so $n_1 = 2$.
• Stage $2$, the flavour: whichever cone we take, chocolate, vanilla or strawberry, so $n_2 = 3$.

The multiplication principle gives

The tree shows the $6$ cones on its bottom row.

Taking the flavour first, the stages swap:

• Stage $1$, the flavour: $n_1 = 3$.
• Stage $2$, the cone: whichever flavour we take, $n_2 = 2$.

The multiplication principle gives

the same $6$ cones in a tree of a different shape.

The multiplication principle only requires each stage to offer the same number of options whatever was chosen before it, not the same options. For example, a bike shop sells a road frame in red, black or silver, and a mountain frame in green, blue or orange. The colours differ between the frames, but each frame offers $3$ of them. In the tree of choices, each frame splits into $3$ leaves, one per bike, so the tree has $2 \times 3 = 6$ leaves and there are $6$ bikes.

Now suppose the mountain frame comes in green or blue only. We are still counting the leaves of the tree, but the road frame splits into $3$ and the mountain frame into $2$, so no single number describes the second stage and there is no product to form. Instead, we count the leaves under each frame and add the counts:

There are $5$ bikes. Whenever the branches in a row split unevenly, we count the leaves under each branch and add them.

A car hire firm offers a hatchback and an estate. The hatchback comes in red, white or grey, and the estate in black, blue or silver. The hatchback can be hired with a manual or an automatic gearbox, but the estate is automatic only. Does the multiplication principle count the possible hires? How many are there?

Solution

We take the stages in turn and ask of each whether it offers the same number of options whatever was chosen before it.

• Stage $1$, the car: hatchback or estate, so $n_1 = 2$.
• Stage $2$, the colour: the colours differ between the cars, but each car offers $3$, so $n_2 = 3$.
• Stage $3$, the gearbox: $2$ for the hatchback but $1$ for the estate, so there is no single $n_3$.

The third stage fails the check, so the multiplication principle does not apply, and $2 \times 3 \times 2 = 12$ is not the number of hires. We count car by car and add. The hatchback gives $3 \times 2 = 6$ hires and the estate gives $3 \times 1 = 3$, so

There are $9$ possible hires.