First note that the notation n! = (n − 1)(n − 2) ... 3 × 2 × 1 so that, for example,
5! = 5 × 4 × 3 × 2 × 1 = 120 .
We call this 'n factorial'.
Worked Examples
In the 8-letter word DOMINATE all the letters are different. Including the one given, how many arrangements are there of these letters?
Show meYou have 8 choices for the first letter, 7 for the next one, etc, giving
| 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 | = 8! |
| = 40 320 arrangements. |
The word NOMINATE also has 8 letters, but two of them are the same. How many arrangements are there now?
Show meAs two of the letters are the same, each arrangement will be repeated. Thus there are
| \frac{8!}{2} | = \frac{40 320}{2} |
| = 20160 arrangements |
The word ADDITION has eight letters with two pairs of identical ones. How many different arrangements are there?
Show meNow we have two pairs of letters repeated, so there are
\frac{8!}{2×2} = 10 080 arrangements
How many arrangements are there of the word CALCULUS?
Show meHere 3 letters are repeated; the number of unique arrangements is
\frac{8!}{2×2×2} = 5040
The word DIVISION has one letter that appears three times. How many arrangements of these eight letters are there?
Show meThere are 6 (= 3!) arrangements of three letters so we are left with
\frac{8!}{3!} = 6720 unique arrangements.
In general, the number of arrangements of an n-letter word with one letter repeated p times is given by the formula
Here, n!, read as 'n factorial', is short for the product of all the natural numbers up to and including n; so for example
7!= 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
The number of different arrangements of letters in the word SIMILAR is therefore
\frac{7!}{2!} = \frac{7 × 6 × 5 × 4 × 3 × 2 × 1}{2 × 1} = 2520
This rule can be extended to cover the repetition of more than one letter. The number of arrangements of letters in the word SENSES is
\frac{6!}{3! 2!} \frac{6 × 5 × 4 × 3 × 2 × 1}{3 × 2 × 1 × 2 × 1} = 60
This automatically provides a way of finding binomial coefficients.
\binom{5}{3} is the number of arrangements of the letters AAADD
so
\binom{5}{3}=\frac{5!}{2!3!}=\frac{5 × 4 × 3 × 2 × 1}{2 × 1 × 3 × 2 × 1}=\frac{5 × 4}{2 × 1}=10
Similarly the binomial coefficient
| \binom{10}{7} | =\frac{10!}{3!7!}=\frac{10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1}{(3 × 2 × 1) × (7 × 6 × 5 × 4 × 3 × 2 × 1)} |
| = \frac{10 × 9 × 8}{3 × 2 × 1} = 120 |
and
\binom{10}{3} = \frac{10!}{3!7!} = 120
In general,
\binom{n}{n-r}=\frac{n!}{(n-(n-r))!(n-r)!}=\frac{n!}{r!(n-r)!}=\binom{n}{r}
Thus the general formula for binomial coefficients is given by
Your calculator probably has a factorial function on it. It may also have a function for working out binomial coefficients; consult your manual if you are not certain. In any case, it is often possible to work them out without a calculator; though the formula looks difficult there is always a good deal of cancelling that can be done, as the examples so far have shown.
How many ways are there of selecting a tennis team of 5 from a squad of 9 players?
Imagine all 9 players in a line, 5 of them are in the team: label them T; 4 are not: label them N. The answer to the question is the number of arrangements of 5 Ts and 4 Ns, since each different arrangement gives a different team.
\binom{9}{5}=\frac{9!}{4!5!}=\frac{9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1}{(4 × 3 × 2 × 1)(5 × 4 × 3 × 2 × 1)}=126
Exercises
