Binomial Coefficients: Selected Exercises

1. Binomial Coefficients: Selected Exercises

2. Preliminaries What is the coefficient of x2y in (x + y)3? (x + y)3 = (x + y)(x + y)(x + y) = (xx + xy + yx + yy)(x + y) = xxx + xyx + yxx + yyx + xxy + xyy + yxy + yyy = x3 + 3x2y + 3xy2 + y3. The answer thus is3. There are 8 terms in the formal expansion. Answer: The # of ways to pick the y position in the formal expansion: C(3, 1).

3. Preliminaries • How many terms are there in the formal expansion of (x + y)n? • How many such terms have exactly 3ys? • This is the coefficient of xn-3y3 in (x + y)n? • How many such terms have exactly jys?

4. The Binomial Theorem Let x & y be variables, and n N. Partition the set of 2n terms of the formal expansion of (x + y)n into n + 1 classes according to the # of ys in the term: (x + y)n = Σj=0 to n C( n, j )xn-jyj= C(n, 0)xny0 + C(n,1)xn-1y1 + … + C(n, j)xn-jyj+ … + C(n,n)x0yn.

5. Pascal’s Identity Let n & k be positive integers, with n > k. Then C(n, k) = C(n - 1, k – 1) + C(n - 1, k). Proof by a combinatorial argument: • The left hand side counts the number of subsets of size k from a set of n elements. • The right hand side counts the same subsets using the sum rule: Partition the subsets into 2 disjoint classes • Subsets of k elements that include element 1: • Pick element 1: 1 • Pick the remaining k – 1 elements from the remaining n - 1 elements: C(n - 1, k – 1). • Subsets of k elements that exclude element 1: • Pick the k elements from the n - 1 remaining elements: C(n - 1, k).

6. *30 Give a combinatorial proof that Σk=1,nkC(n, k)2=nC( 2n – 1, n – 1 ). Hint: Show that the equation’s LHS and RHS are different ways to count the same thing: The # of ways to select a committee with n members from a group of n math professors & n computer science professors, such that the committee chair is a mathematics professor.

7. *30 Solution Proof by a combinatorial argument The RHS counts the # of such committees by: • Pick the chair from the n math professors: n • Pick the remaining n – 1 members from the remaining 2n – 1 professors: C( 2n – 1, n – 1 ) TheLHS counts the same committees by: Partition the set of such committees into disjoint subsets, according to, k, the # of math professors on the committee. For each k, • Pick the k math professor members: C(n, k) • Pick the committee chair: k • Pick the n - k CS professor members: C( n, n – k ) = C( n, k )

8. Combinatorial Identities • Manipulation of the Binomial Theorem • “Committee” arguments • Block walking arguments – for identities involving sums

9. Manipulation of the Binomial Theorem Prove that C(n, 0) + C(n, 1) + . . . + C(n, n) = 2n. In general, • Algebraically manipulate the binomial theorem. • Evaluate the resultant equation for suitable values of x & y to get the desired result. Prove that n2n – 1= 1C(n, 1) + 2C(n, 2) + 3C(n, 3) + . . . + nC(n, n).

10. Committee Arguments Show that • C(n, k)C(k, m) = C(n, m)C(n – m, k – m). Hint: committees of k people, m of whom are leaders. • Σk = 0 to r C(m,k)C(n, r - k) = C(m + n, r). Hint: committees of r people taken from m men & n women.

11. Block-Walking Arguments • Draw Pascal’s triangle. • Interpret a node in the triangle as the # of ways to walk from the apex to the node, always going down. Show that • C(n, k) = C(n – 1, k) + C(n – 1, k – 1) • C(n,0)2 + C(n,1)2 + ... + C(n,n)2 = C(2n,n).

12. Characters •   .≥ ≡ ~ ┌ ┐└┘ •        ≈ •    •  Ω Θ •      Σ ¢ •        

13. *10 Give a formula for the coefficient of xk in the expansion of (x + 1/x)100, where k is an even integer.

14. *10 Solution • By the Binomial Theorem, (x + 1/x)100 = Σj=0 to 100 C(100, j)x100-j(1/x)j = Σj=0 to 100 C(100, j)x100-2j. • We want the coefficient of x100-2j, where k = 100 – 2jj = (100 – k)/2. • The coefficient we seek is C(100, (100 – k)/2).

15. 20 Suppose that k & n are integers with 1  k<n. Prove the hexagon identity C(n - 1, k –1)C(n, k + 1)C(n + 1, k) = C(n-1, k)C(n, k-1)C(n+1, k+1), which relates terms in Pascal’s triangle that form a hexagon. Hint: Use straight algebra.

16. 20 Solution C( n – 1, k –1 )C( n, k + 1 )C( n + 1, k ) = (n – 1)! n! (n+1)! _______________________________________ (k – 1)!(n – k)! (k + 1)!(n – k – 1)! k! (n + 1 – k)! = C( n – 1, k )C( n, k – 1 )C( n + 1, k + 1 ).