(1/16)

MA284 : Discrete MathematicsWeek 4: Permutations and

Combinationshttp://www.maths.nuigalway.ie/˜niall/MA284/

02 October, 20191 Recall...

... Binomial coefficients2 Permutations3 Combinations, again

A formula4 Towards Algebraic and Combinatorial Proofs5 ExercisesThese slides are based on §1.3 and §1.4 of Oscar Levin’sDiscrete Mathematics: an open introduction.They are licensed under CC BY-SA 4.0

(2/16)Announcements

1. Because of Open Day, Friday’s lecture (04 Oct) is cancelled. Sorry!

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2. ASSIGNMENT 1 is now open!To access the assignment, go tohttp://mathswork.nuigalway.ie/webwork2/1920-MA284

Your USERNAME is:Your PASSWORD is:

There are 15 questions.You may attempt each one up to 10 times.This assignment contributes 10% to your final grade for MA284.Deadline: 5pm, Friday 4 October.

Recall... ... Binomial coefficients (3/16)

Binomial CoefficientsFor each integer n ≥ 0, and integer k such that 0 ≤ k ≤ n, there is anumber (

nk

)read as “n choose k”

1.(n

k)

= |Bnk |, the number of n-bit strings of weight k.

2.(n

k)

is the number of subsets of a set of size n each with cardinality k.3.

(nk)

is the number of lattice paths of length n containing k steps tothe right.

4.(n

k)

is the coefficient of xkyn−k in the expansion of (x + y)n.5.

(nk)

is the number of ways to select k objects from a total of nobjects.

Recall... ... Binomial coefficients (4/16)

There are two ways one can compute(

nk

).

(1) The recurrence relation(

nk

)=

(n − 1k − 1

)+

(n − 1

k

)(2) Using the formula,

(nk

)= n!

k!(n − k)! .

We justified the recurrence relation in (1) last week, and used it to makePascal’s Triangle (Which we’ll return to later).

First, we’ll derive (2), by relating it to the number of permutations of nobjects.

Permutations (5/16)

PermutationsA permutation is an arrangement of objects. Changing the order of theobjects gives a different permutation.

ExampleLast Sunday, 7 athletes contested the 100m Women’s final at the WorldAthletics Championship in Doha. How many different finishing orderingswere possible?

Permutations (6/16)

More generally, recall that n! (read “n factorial”) is

n! = n × (n − 1)× (n − 2)× · · · × 2× 1

E.g.,2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 10! = 3, 628, 800.

Number of permutationsThere are

n! = n × (n − 1)× (n − 2)× · · · × 2× 1

(i.e., n factorial) permutations of n (distinct) objects.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

So, there are n! permutations of n objects, where each permutationcontained all n objects. Now consider a more general case.

Permutations (7/16)

Permutations of k objects from nThe number of permutations of k objects out of n, P(n, k), is

P(n, k) = n × (n − 1)× · · · × (n − k + 1) = n!(n − k)!

Example (P(7, 3))Last Sunday, 7 athletes contested the 100m Women’s final at the WorldAthletics Championship. How many different finishing was could thegold, silver, and bronze medals be awarded.

Dina Asher-Smith (GBR), Shelly-Ann Fraser-Pryce (JAM), Marie-Josee Ta Lou (CIV)

Permutations (8/16)

Choosing the half-backs on a rugby team...Ireland Squad for the Men’s Rugby World Cup has 5 half-backs (i.e.,9 and 10): Joey Carbery, Jack Carty, Luke McGrath, Conor Murray, andJohnny Sexton. (Pretend all 5 are competent to play at 9 or 10).

1. How many choices do we have for picking the starting scrum-half (9),out-half (10)?

2. How many choices for picking a starting scrum-half (9), out-half (10),and one substitute (23)?

Combinations, again (9/16)

Combinations (again)A combination is a selection of objects, where order does not matter.That is, it is a set.We have already seen that, if we have a set of n objects, there are

(nk)

subsets of size k.So the number of combinations of k objects out of n is

(nk

).

Now we want to derive the formula for(

nk

).

Combinations, again (10/16)

Choosing the half-backs again...Recall that our rugby team has 5 backs: Carbery, Carty, McGrath,

Murray, and Sexton. There are(

53

)ways we can pick 3 of them for our

match-day squad (i.e., starting team + substitutes).

Once we have picked these three, there are 3! = 6 ways we can assignthem the 9, 10 and 23 jerseys. That is

P(5, 3) =(

53

)3!.

However, we know P(5, 3), so this gives a formula for(5

3).

Combinations, again A formula (11/16)

(1) We know there are P(n, k) permutations of k objects out of n.(2) We know that

P(n, k) = n!(n − k)!

(3) Another way of making a permutation of k objects out of n is to(a) Choose k from n without order. There are

(nk

)ways of doing this.

(b) Then count all the ways of ordering these k objects. There are k!ways of doing this.

(c) By the Multiplicative Principle,

P(n, k) =(

nk

)k!

(4) So now we know that n!(n − k)! =

(nk

)k!

(5) This gives the formula(

nk

)= n!

(n − k)!k!

Towards Algebraic and Combinatorial Proofs (12/16)

On the previous slide, we gave a proof that(nk

)= n!

(n − k)!k!

We did this by counting P(n, k) in two different ways.

This is a classic example of a Combinatorial Proof , where we establish aformula by counting something in 2 different ways.

We’ll now study this style of proof. See also Section 1.4 of the text-book.

But first, we will form some conjectures.

Towards Algebraic and Combinatorial Proofs (13/16)

Binomial coefficients have manyimportant properties.Looking at their arrangement inPascal’s Triangle, several of theseare obvious:

Towards Algebraic and Combinatorial Proofs (14/16)

ProofsProofs of identities involving Binomial coefficients can be classified as

Algebraic: if they rely mainly on the formula for binomialcoefficients.Combinatorical: if the involve counting a set in two different ways.

For our first example, we will give two proofs of the following fact:(nk

)=

(n

n − k

).

Towards Algebraic and Combinatorial Proofs (15/16)

Algebraic proof of Pascal’s triangle recurrence relation(nk

)=

(n − 1k − 1

)+

(n − 1

k

)

Exercises (16/16)

Q1. Put the following numbers in increasing order.

(a) The number of subsets of the set {a, b, c, d , e, g , h, i}.(b)

(105)

(c)(12

3).

(d)(12

3).

(e) 5!(f) P(7, 4)(g) P(8, 5)

Q2. Compute(7

3)

using Pascal’s Identity. Check you got the right answer byalso doing this using the factorial formula.

Q3. Write out all permutations of the letters A, B, C, and D that use all fourletters. Verify you get 24.Now write out all permutations of the 4 letters A, B, C, and C (i.e., C isrepeated). How many do you get?