1. shopping
2. gifts
3. greeting cards

Math 55 Worksheet
Permutations, Combinations
1. Five-card draw is played with a standard, 52-card deck. There are 13 different kinds, or ranks of cards: the
numbers 2 to 10, and the face cards, jack (J), queen (Q), king (K) and ace (A). And there are four cards of each
rank, each in one of the four suits: clubs (♣), diamonds (♦), hearts (♥) and spades (♠). Certain combinations
of cards are called “hands” – these hands are listed below, in order from best to worst. So a player who draws a
four-of-a-kind beats a player who draws a straight. A player’s best hand is the best combination they can make
from their cards. So a player with A♥A♣4♣2♥4♠ can make one pair – aces – but their “best hand” is two pairs,
aces and fours.
How many ways are there for each of these to be drawn as someone’s best five-card hand:
(a) straight flush? (Five consecutive cards of the same suit. The ace can be the highest card in the straight,
or the lowest, but not both at once – A2345 and 10JQKA are allowed, but QKA23 is not.)
(b) four-of-a-kind?
(c) full house? (Three of one kind and two of another.)
(d) flush? (All five cards the same suit.)
(e) straight? (As with a straight flush, the ace may be highest or lowest, but not both.)
(f) three-of-a-kind?
(g) two pair?
(h) one pair?
2. Show that if p is prime and 1 ≤ k < p, then kp is divisible by p. Solution: kp is an integer. And it is also
equal to the a priori “fraction” p!/(k!(p − k)!). Since this is an integer, this must mean that the denominator
must cancel with the numerator. But since the prime factors of the denominator are all less than
p, nothing can
cancel with the prime number p. Hence after cancellation the factor of p survives, so that kp is divisible by p.
3. How many different 5-person committees can be made if there are 100 students and 30 faculty members, if the
committee must contain more students than faculty?
Solution: It’s easiest to just look at cases. If there are 5
students and zero faculty, then there are 100
ways
to
do this. If we choose 4 students and one faculty, then
5
30
30
there are 100
ways
to
do
this.
Finally
if
there
are
3 students and two faculty, there are 100
ways.
1
3
2
4
ANSWER:
100
100 30
100 30
+
+
5
4
1
3
2
4. How many 5 digit, decimal strings can be formed using:
(a) Two 0s and three 3s?
(b) Exactly one 4?
(c) Two 6s and one 7?
Binomial Theorem and Combinatorial Proofs
5. What is the coefficient of x3 y 5 in the expainsion of (2x + 3y)8 Answer: The term with x3 and y 5 is
So the coefficient is
8 3 5
2 3
3
6. Show that
n
X
k=0
8
3
(2x)3 (3y)5 .
n
2
= 3n . Done in Section
k
k
n
2
7. Give a combinatorial proof that 2n
2 = 2 2 + n . Hint: for the LHS, break it down into cases of where the 2
elements come from. We want to count the number of ways of choosing 2 things from a set with 2n objects. Let’s
split up our set of 2n objects into two groups of size n. Call them groups 1 and 2. Now to choose 2 elements of
our original BIG set, we can do this in terms
of the smaller sets in the following ways:
1) Choose 2 elements from group 1 : n2 ways.
2) Choose 2 elements from group 2 : n2 ways.
3) Choose 1 element
from each group. n2 ways.
2n
n
Thus n = 2 2 + n2 .
8. F Compute
n
X
(−1)k
k=0
n
. Then, use this to determine how many odd-sized subsets a set with n elements has.
k
9. F Give a combinatorial proof that
n
r
r
k
=
n
k
n−k
r−k
.
10. F A rectangular city’s roads are laid out like a grid in which (0,0) is the most southwest corner of the city and
(m, n) is the most northeast. If there’s a road at every integer (both north-south and east-west), how many ways
can you get from (0,0) to (m,n) assuming you never backtrack (that is, you never go south or west)?