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 is 3.
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 3 ys?
This is the coefficient of xn-3y3 in (x + y)n?
How many such terms have exactly j ys?

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,n kC(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 )
The LHS 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.
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) C(n,n)2 = C(2n,n).

12. Characters    . ≥ ≡ ~ ┌ ┐ └ ┘        ≈     Ω Θ
   . ≥ ≡ ~ ┌ ┐ └ ┘
       ≈
  
 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)!
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(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 ).