Recall the following basic principle of counting: Suppose a set of objects can be specified by making
a sequence of k choices, and suppose further that the number of possibilities at the ith step is always ci
(independent of the previous choices). Then the total number of objects is c1 c2 · · · ck .
Example 1. How many ways are there to arrange the letters in the word STOP? There are 4 ways to choose
which letter goes first, then, given this choice, 3 ways to choose which letter goes second (since it must be
one of the remaining letters), 2 ways to choose which letter goes third, and 1 way to choose which letter goes
last. Hence the answer is 4 · · · 3 · 2 · 1 = 24.
The same argument shows that the number of ways to order n things is n(n − 1)(n − 2) · · · 2 · 1 = n!, read
n-factorial. By convention, 0! = 1.
Example 2. Six men and six women are to pair up for a mixed doubles tennis tournament. In how many
ways is this possible?
Ans: 6!
Example 3. Five boys and four girls are to line up for a fire drill. In how many ways is this possible if (a)
any order is allowed; (b) all the boys must line up first; (c) the line must alternate boy-girl-boy-girl?
Ans: (a) 9!=362880; (b) 5!4! = 2880; (c) Same as part (b).
Example 4. Mary has 4 math books, 4 history books, and 4 chemistry books. How many ways can she
arrange these on a shelf if books on the same subject must form a contiguous block?
Ans: 4!3 · 3! = 82944.
Example 5. How many ways are there to rearrange the letters in the word BOOK?
While you might think the answer should be 4!, in fact it is only 12. The reason is that since BOOK
has two O’s, we can’t tell which O comes first. In other words, if the O’s were distinguishable, say one
was red and the other blue, then there would be 4! = 24 ways. However, if we ignore the colors, then each
arrangement occurs twice because the O’s can appear in either order, so there are only 12.
Example 6. How many ways are there to rearrange the letters in the word BANANA?
As before, if we could distinguish the N’s and A’s, there would be 6! ways. However, we can’t distinguish
6!
= 60 ways.
the 2! different orders for the N’s, nor the 3! different orders for the A’s, so there are only 2!3!
In general, the number of ways to arrange n objects, of which n1 are of one type, n2 are of the second
type, . . . , and nr are of the rth type is
n!
.
n1 !n2 ! · · · nr !
Example 7. The number of ways to rearrange the letters in the word MISSISSIPPI is
11!
4!4!2!1!
= 34650.
Example 8. How many ways are there to split ten students into a group of size 5, a group of size 3, and a
group of size 2?
10!
Ans: 5!3!2!
= 2520.
Example 9. How many ways are there to choose a committee of three people from a group of ten?
This is the same as the number of ways to split it into a committee of size 3 and a ”non-committee” of
10!
size 7. Hence the answer is 3!7!
.
n!
comes up quite often, so we call it the binomial coefficient nr , read “n-choose-r.”
The quantity r!(n−r)!
As we have just seen, nr is the number of ways to form a group of r objects from an n-element set.
10·9·8
Note that in the previous example, 10
3 =
3! . We could see this, for instance, by choosing one member
of the committee in 10 ways, then another member in 9 ways, and then a third member in 8 ways. But since
we don’t care which order we choose the members in, and there are 3! such orders, we must divide to get
n(n−1)(n−2)···(n−r+1)
10
n
.
3 . In general, r =
r!
1