The Product Rule is used when there is more than one choice from a selection of options. The product rule consists of multiplying the number of selections, at each stage, together:
number of permutations = n1 n2 n3...
where n1 is the number of permutations for the first selection; n2 is the number of permutations for selection 2, and so on.
The number of permutations may change depending on a previous selection (as in a lottery, for example, where a drawn ball cannot be drawn again).
Six coloured stripes are being used as a background to a logo. Red and yellow, the corporate colours, must be included; a further 8 colours are available for the remaining stripes.
How many permutations are there for the background if no colour may be used twice?
You must use red. There are 6 possible positions for red: n1 = 6
You must use yellow. There are 5 remaining positions for yellow: n2 = 5 (one has been used by red)
The remaining colours are optional. you must now use the remaining positions. Choose a position. There are 8 available colours, n3 = 8
Choose the next position, and there are 7 available colours. Repeat for 6 colours and 5 colours on the final two positions: n4 = 7, n5 = 6 and n6 = 5
Using the Product Rule: 6 x 5 x 8 x 7 x 6 x 5 = 50,400 possibilities.
Notice that the first two elements were based on selection of position (as red and yellow are pre-selected), and the last four elements were based on selection of colour.
A random four-letter code is being generated from upper-case letters. A letter cannot be used more than once. The first letter may be any character; the remaining five letters must be consonants. How many combinations are there?
The calculation depends on whether a vowel or a consonant is drawn first, as letters cannot be re-used.
Vowel+consonants: There are 5 vowels that could be in position 1; 21 consonants for position 2; 20 consonants for position 3 and 19 consonants for position 4 giving a total 5 x 21 x 20 x 19 = 39,900 possibilities.
Consonants only: for position 1: 21 possibilities, for position 2 there are 20, for position 3 there are 19 and for position 4 there are 18 = 21 x 20 x 19 x 18 = 143,640 possibilities.
The combined total is 39,900 + 143,640 = 183,540 possibilities.