Combinatorics Worksheet 1
1. How many ways of arranging the letters in BANANA?
2. How many ways are there of writing a 5 letter word:
a. where no two letters are the same?
b. where no two adjacent letters are the same?
3. [SMC] In how many different ways can I circle letters in the grid shown so that there is
exactly one circled letter in each row and exactly one circled letter in each column?
A
D
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
U
V
W
X
Y
15
100
B
E
24
120
C
60
4. The well-known ‘Birthday Paradox’ asks the following question: “How many people do you
need in a room such that there’s about a 50% chance that two people share the same
birthday?”. Show that when there are 23 people, the probability is roughly a half. (Hint:
When determining the probability of ‘at least one thing occurring’, it’s much easier to find
the probability of the event never occurring and subtracting from 1; in this case no one
sharing a birthday)
5. An examiner invigilates an exam in which 1500 students are applying for 150 places at a
school. The examiner is responsible for invigilating a room of 25 students. What is the
chance that all of them get a place? Use:
a. A probabilistic approach (as per the example in the slides).
b. A combinatoric approach.
6. I have two identical packs of cards. I take each pack and shuffle it separately, placing them
both face down in the table. I then proceed to play a game of snap with myself (being a
mathematician I have no friends). I compare the top card from each pile, and then compare
the second card from each pile, and so on. What is the probability that I continue in this way
all the way down the piles, and never find an exact match in the 52 pairs of cards?
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Combinatorics Worksheet 1 - ANSWERS
1. How many ways of arranging the letters in BANANA?
There’s 6 letters, but the ordering of the 3 A’s doesn’t matter and the ordering of the 2 N’s
doesn’t matter. So the number of ways is:
6!
= 60
3! 2!
2. How many ways are there of writing a 5 letter word:
a. where no two letters are the same?
Using a ‘slot filling’ approach, we have: 26P5 = 26 × 25 × 24 × 23 × 22 = 7 893 600
words.
b. where no two adjacent letters are the same?
There’s 26 × 25 × 25 × 25 × 25 = 26 × 254 = 10 156 250 words.
3. [SMC] In how many different ways can I circle letters in the grid shown so that there is exactly
one circled letter in each row and exactly one circled letter in each column?
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
U
V
W
X
Y
In the first row, I can choose any of the 5 letters. Then in the next row, I can only
choose from 4 letters to avoid using the same column. This continues
downwards, giving 5 × 4 × 3 × 2 × 1 = 5! = 𝟏𝟐𝟎 ways.
4. The well-known ‘Birthday Paradox’ asks the following question: “How many people do you
need in a room such that there’s about a 50% chance that two people share the same
birthday?”. Show that when there are 23 people, the probability is roughly a half:
a. A probabilistic approach.
Consider the probability of each person not sharing the same birthday as any of the
previous people considered.
- For the first person, the probability of not clashing is 1.
355
- For the second person, there’s a 356 chance their birthday doesn’t clash with the first.
354
- For the third person, there’s a 356 chance their birthday doesn’t clash with the two
previous people.
Continuing on like this, the probability that 23 people don’t share a birthday is
356!÷(356−23)!
35623
=
356!
333!⋅35623
= 0.484 to 3dp. Thus the probability that at least one person
shares a birthday is 1 − 0.484 = 0.516.
b. A combinatoric approach.
We can avoid clashes by choosing 23 of the days for people to have their birthdays on.
There’s 365P23 such choices if we consider them ordered. But there’s 35623 total possible
choices of birthdays (with or without clashes). The probability is therefore
which will be the same.
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356!/333!
35623
5. An examiner invigilates an exam in which 1500 students are applying for 150 places at a
school. The examiner is responsible for invigilating a room of 25 students. What is the chance
that all of them get a place? Use:
a. A probabilistic approach.
150
The first student has a probability of 1500 of getting in. The next student, if the first got
149
in, has a probability of 1499 of getting in. And so on. The probability is thus
150×149×…×126
1500×1499×…×1476
=
150!×1475!
.
1500×125!
Which is roughly a 1 in 7 × 1025 chance.
b. A combinatoric approach.
In total there’s (1500
) ways of the applicants filling the places. If the 25 students in the
150
room all got a place, then there’s (1475
) ways in which the remaining applicants could fill
125
the remaining places. The probability is therefore
(1500
150 )
(1475
125 )
=
1500!
)
150!1350!
1475!
(
)
125!1350!
(
=
150!×1475!
as
1500×125!
before.
6. I have two identical packs of cards. I take each pack and shuffle it separately, placing them
both face down in the table. I then proceed to play a game of snap with myself (being a
mathematician I have no friends). I compare the top card from each pile, and then compare
the second card from each pile, and so on. What is the probability that I continue in this way
all the way down the piles, and never find an exact match in the 52 pairs of cards?
We have to be careful if we use a probabilistic approach. Suppose we have a simplified pack of 5
cards labelled “a”, “b”, “c”, “d”, “e”. Suppose the first pack is in this order. Suppose on the
second pack I’ve turned over “b” and “c” so far. Then on the third turnover, then the probability
I don’t ‘snap’ is 1 because we’ve already used up the “c” in the second pack. But if in the second
2
pack we’d turned over “d” and “e” so far, then there’s a probability that the third card doesn’t
3
snap. So a probabilistic approach somewhat goes down the toilet, as we’d have a badass
probability tree to contend with.
We’ll have to use a combinatoric approach instead! Suppose we fix the order of the first pack.
Then for the second pack, there’s 52! total possible orderings. To get orderings in the second
pack such that we won’t have a ‘snap’, there’s 51 possibilities for the first card, 50 for the
second, and so on, until we have 1 choice for the penultimate card and similarly 1 choice for the
51!
1
last card. That gives a probability of 52! = 52.
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