Combination Formula Proof - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics

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CSci 2011

Discrete Mathematics

Lecture 21

Combination formula proof

Note that the textbook explains it slightly

differently, but it is same proof

r n r

n

r r r

n n r

P r r

P n r

C n r

Examples

 How many bit strings of length 10 contain:

 at least four 1’s?

 There can be 4, 5, 6, 7, 8, 9, or 10 occurrences of 1  Thus, the answer is:  C (10,4) + C (10,5) + C (10,6) + C (10,7) + C (10,8) + C (10,9) + C (10,10) = 210+252+210+120+45+10+ = 848  Alternative answer: subtract from 2 10 the number of strings with 0, 1, 2, or 3 occurrences of 1

 an equal number of 1’s and 0’s?

 Thus, there must be five 0’s and five 1’s  Find the positions of the five 1’s  Thus, the answer is C (10,5) = 252

Corollary 1

Let n and r be non-negative integers with

r ≤ n. Then C ( n , r ) = C ( n , n-r )

Proof:

Combinatorial proof

A combinatorial proof is a proof that uses counting arguments to prove a theorem  Rather than some other method such as algebraic techniques Essentially, show that both sides of the proof manage to count the same objects Most of the questions in this section are phrased as, “find out how many possibilities there are if …”  Instead, we could phrase each question as a theorem:  “Prove there are x possibilities if …”  The same answer could be modified to be a combinatorial proof to the theorem

Example

 How many ways are there to sit 6 people around a circular table, where seatings are considered to be the same if they can be obtained from each other by rotating the table?  First, place the first person in the north-most chair  Only one possibility  Then place the other 5 people  There are P(5,5) = 5! = 120 ways to do that  By the product rule, we get 1*120 =  Alternative means to answer this:  There are P(6,6)=720 ways to seat the 6 people around the table  For each seating, there are 6 “rotations” of the seating  Thus, the final answer is 720/6 = 120

Binomial Coefficients

It allows us to do a quick expansion of ( x + y )

n

Why it’s really important:

It provides a good context to present proofs

 Especially combinatorial proofs

Let n and r be non-negative integers with

r ≤ n. Then C ( n , r ) = C ( n , n-r )

Proof (from last slide set):

Review

Polynomial expansion

Consider ( x + y )

3

Rephrase it as: (x+y)

3

= x

3

+ 3 x

2

y + 3 xy

2

+ y

3

(x+y)(x+y)(x+y) = x

3

+ x

2

y + x

2

y + x

2

y + xy

2

xy

2

+ xy

2

+ y

3

When choosing x twice and y once, there are C(3,2)

= C(3,1) = 3 ways to choose where the x comes

from

When choosing x once and y twice, there are C(3,2)

= C(3,1) = 3 ways to choose where the y comes

from

Polynomial expansion

Consider: (x+y)

5

=x

5

+5x

4

y+10x

3

y

2

+10x

2

y

3

+5xy

4

+y

5

To obtain the x

5

term

 Each time you multiple by ( x + y ), you select the x  Thus, of the 5 choices, you choose x 5 times: C(5,5) = 1  Alternatively, you choose y 0 times: C(5,0) = 1

To obtain the x

4

y term

 Four of the times you multiply by ( x + y ), you select the x  The other time you select the y  Thus, of the 5 choices, you choose x 4 times: C(5,4) = 5  Alternatively, you choose y 1 time: C(5,1) = 5

To obtain the x

3

y

2

term

 C(5,3) = C(5,2) = 10

Pascal’s triangle

n =

Pascal’s Identity

By Pascal’s identity: 7 C 5 = 6 C 5 + 6 C 4 or 21=15+

Let n and k be positive integers with n ≥ k.

Then

n+

C

k

n

C

k

n

C

k-

We will prove this via two ways:

 Combinatorial proof  Using the formula for (^) nC (^) k

Algebraic proof of Pascal’s identity ( 1 ) ( 1 )( )!

n k n k n k n n n n k n k n k − + = − + −

n + 1 = n + 1

k n k n k n k n k n k n −

k k n k n k n k n k n k k n k n k n n − −

k n k n k k n 1 ( 1 )

n + 1 = k + n − k + 1

k n k n k k n k k k n k n

Substitutions:

Pascal’s triangle

0 1 2 3 4 5 6 7 8 n = 1 2 4 8 16 32 64 128 256 sum = = 2 n