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Lecture 7
1. Unordered Selection, continued
Let us recall the following:
Theorem 7.1. The number of ways to create a team of r things among n
is “n choose r.” Its numerical value is
( n
r
n!
r!(n − r)!
Example 7.2. If there are n people in a room, then they can shake hands in ( n 2
many different ways. Indeed, the number of possible hand shakes is the
same as the number of ways we can list all pairs of people, which is clearly ( n 2
. Here is another, equivalent, interpretation. If there are n vertices in
a “graph,” then there are
n 2
many different possible “edges” that can be
formed between distinct vertices. The reasoning is the same.
Example 7.3 (Recap). There are
52 5
many distinct poker hands.
Example 7.4 (Poker). The number of different “pairs” [a, a, b, c, d] is
choose the a
×
deal the two a’s
×
choose the b, c, and d
× 4
3 ︸︷︷︸ deal b, c, d
Therefore,
P(pairs) =
13 ×
4 2
×
12 3
× 4
3 ( 52 5
Example 7.5 (Poker). Let A denote the event that we get two pairs
[a, a, b, b, c]. Then,
|A| =
choose a, b
×
deal the a, b
× 13
choose c
deal c
Therefore,
P(two pairs) =
13 2
×
4 2
× 13 × 4
52 5
Example 7.6. How many subsets does { 1 ,... , n} have? Assign to each
element of { 1 ,... , n} a zero [“not in the subset”] or a one [“in the subset”].
Thus, the number of subsets of a set with n distinct elements is 2 n .
Example 7.7. Choose and fix an integer r ∈ { 0 ,... , n}. The number of
subsets of { 1 ,... , n} that have size r is
n r
. This, and the preceding proves
the following amusing combinatorial identity:
∑^ n
j=
n
j
n .
You may need to also recall the first principle of counting.
The preceding example has a powerful generalization.
Theorem 7.8 (The binomial theorem). For all integers n ≥ 0 and all real
numbers x and y,
(x + y) n =
n ∑
j=
n
j
x j y n−j .
Remark 7.9. When n = 2, this yields the familiar algebraic identity
(x + y) 2 = x 2
For n = 3 we obtain
(x + y) 3 =
x 0 y 3
x 1 y 2
x 2 y 1
x 3 y 0
= y
3
2
2 y + x
3 .
