Counting, Permutations & the Binomial Theorem — Challenge
Expanding a bracket and counting arrangements turn out to be the same question.
What these problems cover
These questions use 4 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Independent — The first event does not change the second — like rolling a die twice.
- Coefficient — The number multiplying a variable — the 3 in 3n.
- Permutation — An arrangement where order matters. P(n,r) = n!(n-r)!.
- Combination — A selection where order does not matter. C(n,r) = n!r!(n-r)!.
Problem 1
In how many different orders can 4 distinct books be arranged on a shelf?
- 4
- 12
- 16
- 24
- 256
Worked answer
24
There are 4 choices for the first position, then 3 for the next, then 2, then 1.
Multiplying gives 4· 3· 2· 1 = 24, which is 4!.
44 = 256 would be the count if repeats were allowed, which they are not here.
Problem 2
In how many ways can 2 items be chosen from 5, when order does not matter?
- 5
- 10
- 20
- 25
- 120
Worked answer
10
Choosing 2 from 5 in order gives 5· 4 = 20 possibilities.
Order does not matter here, and each pair has been counted twice, once in each order.
So the answer is 202 = 10.
Problem 3
How many possible outcomes are there when three coins are flipped?
- 3
- 6
- 8
- 9
- 27
Worked answer
8
Each coin has 2 outcomes, and the coins are independent.
So the total is 2· 2· 2 = 23 = 8.
Listing them gives HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
Problem 4
What is 5!?
- 20
- 25
- 100
- 120
- 720
Worked answer
120
A factorial multiplies all the whole numbers from 1 up to that number.
5! = 5· 4· 3· 2· 1 = 120.
720 is 6!, which is the next one up.
Problem 5
A fair die is rolled twice. What is the probability of getting at least one six?
- 13
- 1136
- 136
- 2536
- 16
Worked answer
1136
Work with the opposite event: no six at all. Each roll avoids a six with probability 56.
So no six in two rolls has probability 2536.
At least one six is everything else: 1-2536 = 1136. Adding 16+16 would double-count the case of two sixes.
Problem 6
How many three-letter codes can be made from the 26 letters if letters may repeat?
- 78
- 2600
- 15600
- 17576
- 26
Worked answer
17576
Each of the three positions can be filled in 26 ways, independently of the others.
26· 26· 26 = 263 = 17576.
If repeats were banned the count would be 26· 25· 24 = 15600, which is the other large option.
Problem 7
In how many ways can 3 items be chosen from 6, when order does not matter?
- 6
- 18
- 20
- 120
- 216
Worked answer
20
Choosing 3 from 6 in order gives 6· 5· 4 = 120.
Each group of three has been counted 3! = 6 times, once per ordering.
So the answer is 1206 = 20.
Problem 8
In how many orders can 3 people be seated in a row?
- 3
- 6
- 9
- 27
- 1
Worked answer
6
There are 3 choices for the first seat, 2 for the second and 1 for the last.
3· 2· 1 = 6 orders.
Problem 9
Two fair dice are rolled. What is the probability that the total is 2?
- 136
- 118
- 112
- 16
- 19
Worked answer
136
There are 36 equally likely ordered outcomes.
A total of 2 needs both dice to show 1, which is exactly one outcome.
So the probability is 136, the smallest possible total and the least likely.