Binomial Coefficients MATH Benjamin V.C. Collins, James A. Swenson MATH 2730

Transcription

1 MATH 2730 Benjamin V.C. Collins James A. Swenson

2 Binomial coefficients count subsets Definition Suppose A = n. The number of k-element subsets in A is a binomial coefficient, denoted by ( n k or n C k or C(n, k, and pronounced n choose k. Example There are six 2-element subsets in any 4-element set, so ( 4 2 = 6. In {a, b, c, d}, they are {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}.

3 First examples Example If n N: ( n 0 = ( n n = ( n = n (n ( n n = n (n

4 Triangular numbers Proposition Let n Z. If n 2, then ( n n = t. 2 t=0 Proof. We ask: How many 2-element subsets are there in {, 2,..., n}? By definition, the answer is ( n 2. On the other hand, there are: 0 subsets whose greatest element is, subset whose greatest element is 2,. t subsets whose greatest element is t +,. n subsets whose greatest element is n. n In total, this makes (n = t subsets. t=0

5 Formula for binomial coefficients Theorem If n, k N and k n, then ( n k = n! k!(n k!.

6 Formula for binomial coefficients Theorem If n, k N and k n, then ( n k = n! k!(n k!. Combinatorial proof. Let A be a set of n elements. We ask: How many k-element subsets are there in A? By definition, the answer is ( n k. On the other hand, let P denote the set of orderings of A; thus P = n!. An element of P can be used to define a k-element subset: just pick the first k elements. We say two orderings are equivalent provided they have the same first k elements, not necessarily in the same order. This is an equivalence relation, and each equivalence class represents a unique k-element subset. Each equivalence class contains k!(n k! permutations, by the n! Multiplication Principle, so there are k!(n k! k-element subsets in A.

7 Exam strategies Example Drew must answer five of the eight questions on a certain exam. In how many ways can Drew choose which questions to answer? What if Drew is required to answer the first two questions?

8 Copying strategies Example The Duplicating Center has eight high-speed copiers, and seven employees who can operate them. There are four identical jobs to be done simultaneously [4000 booklets must be made, in four boxes of 000]. How many ways are there to assign these jobs to operators and machines? Solution. There are ( ( 7 4 ways to choose the employees to do the job, and 8 4 ways to choose machines. Once these choices are made, there are 4! ways to assign operators to machines. Thus there are 4! ( 7 8 4( 4 ways to get the job done, by the Multiplication Principle.

9 Counting bitstrings Exercise A bitstring is a list of 0s and s. How many bitstrings of length n are there? How many bitstrings of length n contain exactly one zero? How many bitstrings of length n contain exactly two zeros? How many bitstrings of length n contain exactly k zeros? Prove: 2 n = n k=0 ( n. k

10 Animal committees Example Seven aardvarks, five giraffes, and three lions are on the advisory board at the Henry Virus Zoo. How many ways are there to select a committee of five animals? This means selecting a 5-element subset from the board, which is a set of = 5 animals. By definition, there are ( 5 5 ( of these. 5 5 = 5! 5!0! = ! = = ! 0! !

11 Animal committees Example Seven aardvarks, five giraffes, and three lions are on the advisory board at the Henry Virus Zoo. How many ways are there to select a committee of exactly three aardvarks and two giraffes? This means selecting 3 of the 7 aardvarks and 2 of the 5 giraffes; by the Multiplication Principle, there are ( 7 5 3( 2 choices. ( 7 ( = 7!5! 3!4!2!3! = !2! = = 350.

12 Animal committees Example Seven aardvarks, five giraffes, and three lions are on the advisory board at the Henry Virus Zoo. How many ways are there to select a 5-member committee with at least one lion? First, count the ones that don t have at least one lion! There are ( 2 5 of these. That leaves ( ( that have at least one lion; this is = 22. Another ( way: 3 ( 2 ( ( 2 ( ( = = 22. Someone might say: ( 3 4 ( 4 = 3003 by the Multiplication Principle: first choose one of the three lions, then choose any four of the remaining animals. Why doesn t this work? [The fact that ( 3 4 ( ( 4 = 5 5 is a coincidence.]

13 Animal committees Example Seven aardvarks, five giraffes, and three lions are on the advisory board at the Henry Virus Zoo. How many ways are there to select a 5-member committee with at least one lion and one giraffe? ( First, count the ones that have no lions. There are 2 5 of these. Next, count the ones that have no giraffes. There are ( 0 5 of these. Oops! Both times, we counted the ones that have neither a lion nor a giraffe! There are ( 7 5 of these. In all, there are ( ( ( 5 0 ( committees with at least one lion and one giraffe.

14 Animal committees Example Seven aardvarks, five giraffes, and three lions are on the advisory board at the Henry Virus Zoo.

15 How many are there? Theorem (principle of inclusion and exclusion If A and B are finite sets, then A + B = A B + A B. Corollary If A is the set of committees with no lions and B is the set of committees with no giraffes, then the number of committees with no lions or no giraffes is A B = A + B A B.

16 Animal committees Example Seven aardvarks, five giraffes, and three lions are on the advisory board at the Henry Virus Zoo. How many ways are there to select a 5-member committee with ( at ( least one aardvark, one giraffe, and one lion? ( 5 0 ( 5 8 ( ( ( ( 5 0 5

17 Have you seen this before? Pascal s triangle

18 Pascal s triangle ( 6 0 ( 0 ( 0 ( ( 0 2 ( 2 ( 2 ( ( 3 ( 3 ( 3 ( ( 4 ( 4 ( 4 ( 4 ( ( 5 ( 5 ( 5 ( 5 ( 5 0 ( ( 6 ( 6 ( 6 ( 6 ( Pascal s triangle

19 Pascal s triangle ( 6 0 ( 0 ( 0 ( ( 0 2 ( 2 ( 2 ( ( 3 ( 3 ( 3 ( ( 4 ( 4 ( 4 ( 4 ( ( 5 ( 5 ( 5 ( 5 ( 5 0 ( ( 6 ( 6 ( 6 ( 6 ( Pascal s triangle

20 Pascal s triangle ( ( 6 ( 6 ( ( 6 Using Pascal s identity

21 Symmetry Proposition If n, k N, then ( ( n k = n Example n k. ( 6 2 ( 6 4 = = 6! 2!(6 2! = = 5; 6! 4!(6 4! = = 5.

22 Symmetry Proposition If n, k N, then ( ( n k = n Algebraic proof. By our formula, ( n = n k n k. n! (n k!(n (n k! = n! (n k!k! = n! k!(n k! = ( n. k

23 Symmetry Proposition If n, k N, then ( ( n k = n n k. 2-elt. subsets 4-elt. subsets 2-elt. subsets 4-elt. subsets {, 2} {3, 4, 5, 6} {2, 6} {, 3, 4, 5} {, 3} {2, 4, 5, 6} {3, 4} {, 2, 5, 6} {, 4} {2, 3, 5, 6} {3, 5} {, 2, 4, 6} {, 5} {2, 3, 4, 6} {3, 6} {, 2, 4, 5} {, 6} {2, 3, 4, 5} {4, 5} {, 2, 3, 6} {2, 3} {, 4, 5, 6} {4, 6} {, 2, 3, 5} {2, 4} {, 3, 5, 6} {5, 6} {, 2, 3, 4} {2, 5} {, 3, 4, 6}

24 Symmetry Proposition If n, k N, then ( ( n k = n n k. Combinatorial proof. Let 0 k n, and let A be a universal set of n elements. For every k-element subset B A, the complement B is an (n k-element subset. Since every k-element subset has a unique complement, the number of k-element subsets in A equals the number of (n k-element subsets in A, which is what we wanted to prove.

25 Pascal s identity Theorem If n, k N and n, then ( ( n k = n ( k + n k. Blaise Pascal (

26 Pascal s identity Theorem If n, k N and n, then ( ( n k = n ( k + n k. Algebraic proof. By our formula, ( n ( n + k k = = = = (n! (k!((n (k! + (n! (k!((n k! k(n! (n k(n! + (k!(n k! (k!(n k! (n!(k + (n k (k!(n k! n! ( n (k!(n k! =. k

27 Pascal s identity Theorem If n, k N and n, then ( ( n k = n ( k + n k. Combinatorial proof. Let 0 k n, let A be a set of n elements, and let w A. We ask: How many k-element subsets are there in A? By definition, the answer is ( n k. On the other hand, ( n k is the number of k-element subsets in A that contain w, because this is the number of (k -element subsets in A \ {w}. Likewise, ( n k is the number of k-element subsets in A that do not contain w. Thus ( ( n k + n k is the answer to our question.

28 Example: combinatorial proof Exercise Prove: ( ( 2n+2 n+ = 2n ( n n n + ( 2n n. Combinatorial proof. We ask: From a group of 2n aardvarks and 2 lions, in how many ways can we form a committee of n + animals? By the definition of binomial coefficient, the answer is ( 2n+2 n+. On the other hand, let s call the lions Leo and Lola. There are ( 2n n committees that include Leo but not Lola, and ( 2n n others that include Lola but not Leo. There are also ( 2n n committees that include both lions, and ( 2n n+ committees that include neither. All told, there are ( ( 2n n n ( n + 2n n possible committees.

29 Why are they called binomial coefficients? (x + y 0 = (x + y = x + y (x + y 2 = x 2 + 2xy + y 2 (x + y 3 = x 3 + 3x 2 y + 3xy 2 + y 3 (x + y 4 = x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4 Theorem (binomial theorem n If n N, then (x + y n =. k=0 ( n k x n k y k.

30 Why are they called binomial coefficients? Theorem (binomial theorem n If n N, then (x + y n = k=0 ( n k x n k y k. Exercise Find the coefficient of x 0 y 3 in (x y 3. Solution. By the binomial theorem, 3 ( 3 (x y 3 = x 3 k ( y k = k k=0 3 k=0 ( 3 ( k x 3 k y k. k So the coefficient of x 0 y 3 is ( 3( 3 3 = 3! 3!0! = = 286.