Counting

Question 1

Define Permutations and how to work them out.

Answer 1

A permutation of n distinct items is any arrangement of them in a specific order. The number of ways to arrange n distinct items is: n! (For permutations of a whole set… Like permutations of {1,2,3,4})

If we are permuting r items out of n distinct items (where r ≤ n), the number of permutations is given by: n!/(n-r)! (For permutations of a subsetset… Like selecting permutations of 3 from 10. n=10 and r=3)

(Ordered selections without repeats)

Question 2

How many Permutations of {1,2,3,4}?

Answer 2

4!

4x3x2x1=24

Question 3

Select 3 Items from 10 items:
Ordered selections without repeats.

Answer 3

10 x 9 x 8 = 720

or

10!/(10-3)! = 720

Ordered selections without repeats. So Permutation

Question 4

Select 3 Items from 10 items:
Ordered selections with repeats.

Answer 4

10³ = 10 x 10 x 10 = 1000

Question 5

Select 3 Items from 10 items:
Unordered selections without repeats.

Answer 5

n choose r

(10 x 9 x 8) / (3 x 2 x 1)

n!/r!(n-r)!

Question 6

How many distict arrangements of the word BARBARA are there?

Answer 6

B = 2, A = 3, R = 2, Length = 7

7!/(2!3!2!)

Or, 7C2 x 5C3 (7C2 choices for B’s, 5C3 choices for A’s, R’s go in whats left)

Question 6

How many ordered selections of 4 elements from {1,2,…,10} have a 5 immediately followed by a 6.

What is the process, what do you use?

Answer 6

Treat 56 as a block. 56 can be arranged in a group of 4, 3 ways.
5 6 _ _ , _ 5 6 _ , _ _ 5 6 (This counts as 3x)

Now there are 2 numbers left to choose and 8 numbers to choose from. (8x)

Finally 1 number left to choose and 7 to choose from. (7x)

Hence, 3x7x8

Can also be done as 8C2 x 3!

Question 7

If a committee of 5 is chosen at random from a group of 25 mathematicians and 25 physicists, what is the probability that at least one mathematician will be chosen?

Answer 7

• The total number of 5-element subsets of the group is 50C5.
• The number of 5-element subsets that contain zero mathematicians is 25C5
• So the number of 5-element subsets that contain at least one mathematician is 50C5 − 25C5
• So the probability is …

(50C5 − 25C5)/50C5

The probability is equal to the number of 5-element subsets of the group that contain at least one mathematician, divided by the total number of 5-element subsets.

Question 8

How to work out probability

Answer 8

P(E) = |E|/|S|

E = No. of success, S = Total attempt

E = No. of success, S = Total attempt

Question 9

What is the coefficient of 𝑥³ in the expansion of (𝑥 + 9)⁷

Answer 9

9^4 x 7C4

Question 10

What is the coefficient of 𝑥² in the expansion of (3𝑥 − 2)⁵

Answer 10

5C3 x 3² x (-2)³

Question 11

State the Inclusion/Exclusion General Rule?

What’s it do?

Answer 11

|𝐴₁ ∪ 𝐴₂| = |𝐴₁| + |𝐴₂| − |𝐴₁ ∩ 𝐴₂|

Stops the intersection of the union being counted twice.

Always helps to draw a Venn diagram

Question 12

In a list of 100 words, there are 47 that contain an “a” and 18 that contain a “b”. If there are 51 that contain neither an “a” nor a “b”, then how many contain both an “a” and a “b”?

Set it up!

Answer 12

|A| = 47, |B| = 18, |A ∪ B| =100 - 51= 49

49 = 47 +18 - |A ∩ B|
=> |A ∩ B| = 47 + 18 - 49
=> |A ∩ B| = 16

Question 13

What is the Pigeonhole Principle.

Also, what is it’s general form?

Answer 13

Pigeonhole Principle

If n items are placed into m pigeonholes and n > m, then at least one pigeonhole must contain more than one item.

In a more general form: If n items are distributed among m pigeonholes, and n > km), then at least one pigeonhole contains more than k items.

Question 14

How many functions are there from Z₅ to Z₃?

Answer 14

Mapping to Z₃, so there is 3 spaces, each with 5 choices:

3^5 = 243