Pigeon-hole principle
2.3. PERMUTATIONS AND COMBINATIONS 69 Example 6 (Binomial coefficients)
Find the number of triangles that can be formed in a convex poly- gon of n sides.
Solution: A polygon is convex if the segment joining any two points on or inside the polygon lies entirely in the polygon. Intu- itively, a convex set is one in which we cannot playhide and seek.
Let the vertex set of the polygon beV ={1,2, . . . , n}. Note that in a convex polygon no three vertices are collinear. Hence a trian- gle is obtained by choosing/joining any three vertices of the setV.
But by the definition of the binomial coefficient a three-element subset of an n-set can be chosen in (n
3
) ways. Hence the number of triangles formed is (n
3
)=n(n−1)(n−2)/3!
Example 2.3.7 (Binomial coefficients):
Find the number of diagonals of a polygon of n sides.
Solution: Let the vertex set of the polygon be V = {1,2, . . . , n}. By joining any two distinct vertices of the polygon we get either a side of the polygon which can be any segment of the set S={12,23,34, . . . ,(n−1)n, n1}or else a diagonal of the polygon. But the number of line segments that can be obtained by joining/choosing any two vertices is the same as the number of two-element subsets of the vertex set V which is the binomial coefficient (n
2
) = n(n−1)/2. Among these n(n −1)/2 segments, exactly n line segments are the sides of the polygon. Hence the number of diagonals =n(n−1)/2−n =n(n−3)/2.
Definition 2.3.1 (Extension of binomial coefficients):
The formula ofTheorem 2.3.2, may be used to define binomial co- efficients(n
k
) whenn isnot a non-negative integer. More precisely, for any real number x, and any non-negative integerk, we define
(x k
)
= x(x−1)· · ·(x−k+ 1) k(k−1). . .(1) = xk
k!
(x k
)
= 0 ifk <0
Corollary 2.3.2.1:
The product of k consecutive positive integers is divisible by k!.
Proof. Consider anyk successive positive integersn+ 1, n+ 2,· · ·, n+k for n ≥ 0. We have to prove that the product (n+ 1) × (n + 2) × · · · × ×(n +k) is divisible by k!. In other words, it is to be proved that (n+1)×(n+2)k!×···××(n+k) is an integer. But this quotient is equal to the binomial coefficient (n+k
k
)(by Theorem 2.3.2.) But the binomial coefficients are integers by definition.
Hence the corollary.
A positive integer p ≥ 2 is called a prime number if its only positive divisors are unity and itself.
Corollary 2.3.2.2:
Ifp is a prime number, then the binomial coefficients (p
1 )
, (p
2 )
, . . . , ( p
p−1 )
are all divisible by p.
Proof. Let k be an integer such that 1 ≤ k ≤ p−1. We have to show that p divides (p
k
). By Corollary 2.3.2.1, k! divides the productp(p−1)· · ·(p−k+ 1). In other words, the quotient
p× (p−1)(p−2)· · ·(p−k+ 1) k!
is an integer. Sincepis prime andk < p, we have,k, k−1, . . . ,2,1 are all relatively prime to p. In particular, k! is relatively prime to p. Therefore in the above quotient, k! divides the product (p − 1)(p − 2)· · ·(p − k + 1) (since if c divides ab and if a and c are relatively prime, then the integer c divides the in- teger b.). That is, (p−1)(p−2)k!···(p−k+1) is an integer. But then, (p
k
)/p= (p−1)(p−2)···(p−k+1)
k! Hence the corollary.
Example 2.3.8 (Duality relation: Binomial coefficients): (n
0
)=n!/0!(n−0)! = 1(since 0! = 1). Similarly, (n
n
) = 1. ( n
n−k
) =
2.3. PERMUTATIONS AND COMBINATIONS 71 n!/(n−k)!(n−(n−k))! =n!/k!(n−k)! =(n
k
) Thus we have the duality relation of the binomial coefficients: (n
k
) =( n
n−k
). There is a nice combinatorial interpretation of this duality relation: every time we select ak-subset of ann-set, we indirectly reject an (n−k)- subset (the complement of the k-subset we select) of then-set.
Example 2.3.9:
If m horizontal parallel lines are cut by n vertical parallel lines, find the number of parallelograms that can be obtained.
Solution: To form a parallelogram, we need two horizontal parallel lines and two vertical parallel lines. Two horizontal lines can be chosen from among the m lines in (m
2
) ways and similarly two vertical lines can be chosen fromnlines in(n
2
)ways. Hence by the product rule (Proposition 2.2.2), the number of parallelograms obtained is (m
2
)×(n
2
)= m(m2−1) × n(n2−1) = mn(m−41)(n−1).
The number of k- multisubsets of an n-set:
The principal utility of the notion of a set is the following: Given an element, we should be able to say if the given element belongs to the set or not. So, in a set, the occurrence of an element isnot repeated more than once.
If we allow the occurrence of an element more than once, the set is called amultisetand the number of occurrences of an element in a multiset is called its multiplicity.
Consider a finite set A. A set B is called a multisubset of A, if every element of B is in the set A and an element of B may occur more than once inB. If the multisubset B ofA has exactly k elements (multiplicity of each element of B is counted), then B is a k-multiset of A. A set can be viewed as a multiset with the multiplicity of each element 1.
Example 2.3.10 (Multisubset):
LetA={1,2,3,4}.Then B ={1,1,1,3,4,4} is a 6-multisubset of A. The multiplicity of the element 1 is 3, of the element 3 is 1, and of the element 4 is 2. Another multiset (in fact simply a subset) ofA is {1,3} where the multiplicity of each element is 1.
We are interested in finding a formula for the number of k- multisubsets of an n-set. We employ a technique called bijective proof. In this proof, we establish a one-to-one correspondence between the set of allk-multisubsets of an n-set and the set of all k-subsets of an n+k−1-set.
Theorem 2.3.3:
The number of k-multisubsets of an n-set is (n+k−1
k
).
Proof. (Bijective proof) Consider an n-set A = {1,2, . . . , n}. A k-multisubset B of A can be written in the form
B ={b1, b2, . . . , bk}
where thebielements of the setB satisfy the following inequalities:
1≤b1 ≤b2 ≤b3· · · ≤bk ≤n
Note that if the strict inequality holds everywhere then the set B becomes simply ak-subset of A. Consider then+k−1-set
A′ ={1,2, . . . , n+k−1}
We shall now establish a one-to-one correspondence between the k-multisubsets of A and k-subsets of A′.
Given a k-multisubset B of the n-set A, we shall produce ak- subsetB′of then+k−1-setA′ by usingB.LetB ={b1, b2, . . . , bk} be ak-multisubset of A with
1≤b1 ≤b2 ≤ · · · ≤bi ≤ · · · ≤bk ≤n
Now the set B′ = {b1 + 0, b2 + 1, . . . , bi+i−1, . . . , bk +k−1} is certainly a k-multisubset of A′, because 1 ≤ bi ≤ n and hence 1≤bi+i−1≤n+i−1≤n+k−1 for all i with 1≤i≤k. We now show that B′ is a k-subset of A′, that is, the elements of B′ aredistinct. If not, supposebi+i−1 =bj+j−1 withi < j.Since i < j, we havebi ≤bj.But then the equationbi+i−1 = bj+j−1 impliesbi−bj =j−i. This is impossible since the left-hand side of this equality bi−bj is ≤0 while the expression on the right-hand side is >0. Hence B′ is a k-subset of A′.
2.3. PERMUTATIONS AND COMBINATIONS 73 Conversely, consider now a k-subsetB′ ={b1, b2, . . . , bk}of A′ with
1≤b1 < b2 <· · ·< bi· · ·< bk ≤n+k−1.
We shall construct a k-multisubset B of A with the help of the k-subset B′.Define B′ ={b1−0, b2 −1, . . . , bi−(i−1), . . . , bk− (k −1)}. We shall prove that B′ is a k-multisubset of A. It is enough if we prove 1≤bi−(i−1)≤n for all i with 1≤i≤k.
First we prove that bi−(i−1)≥1 for all iwith 1≤i≤k by induction oni.This is clearly true fori= 1 forb1−(1−1) =b1 ≥1.
Supposebp−(p−1)≥1 with 1≤p < k.We shall prove it forp+1.
That is, we have to prove bp+1−(p+ 1−1)≥1. Since bp < bp+1, we have bp −p < bp+1 −p and hence bp −p+ 1 ≤ bp+1 −p. By induction hypothesis,bp−p+ 1≥1. Hence bp+1−p≥1.
Next we prove bi−(i−1) ≤ n for all i with 1 ≤ i ≤ k. This is true for i = k for bk ≤ n +k −1 and hence bk−k + 1 ≤ n.
Now bk−1 < bk. Hence bk−1 − k < bk − k. This implies that bk−1−k+ 1≤bk−k and thereforebk−1−k+ 2≤bk−k+ 1≤n.
Hence the inequality is proved for i = k −1. Similarly we prove the inequality successively for i=k−2, k−3, . . . ,1.
Hence we have established a bijection from the set of all k-multisubsets of then-setAonto thek-subsets of the (n+k−1)- set A′. But the number of k-subsets of the (n + k − 1)-set is (n+k−1
k
) by definition. Hence by Proposition 2.2.1, the number of k-multisubsets of the n-set is also equal to (n+k−1
k
).
Example 2.3.11 (Number of multisubsets):
The number of 2-multisubsets of the 3-set {1,2,3} is (3+2−1
2
) = (4
2
) = 4×3/2! = 6.These 2-multisubsets are{1,2},{1,3},{2,3}, {1,1},{2,2},{3,3}.
Example 2.3.12:
If a set of k-dice are thrown simultaneously, find the number of possible outcomes.
Solution: The faces of each die are marked with the integers 1,2,3,4,5,6. Hence an outcome of a single throw of a set of k-dice may be viewed as ak-multisubset of the 6-set{1,2,3,4,5,6}(e.g.,
possible outcomes are
k 1′s
z }| { {1,1, . . . ,1},
(k−1) 1′s
z }| {
{1,1, . . . ,1,2},etc.). Hence the number of possible outcomes is (Theorem 2.3.3) (6+k−1
k
) = (k+5
k
) = (k+5
5
) by the duality relation of the binomial coefficients (Example 2.3.8).
An interpretation of the binomial coefficients as the number of nonnegative integral solutions of a homogeneous linear equation:
Consider the following equation:
x1+x2 +· · ·+xn=k
where k is a positive integer. How many nonnegative integral solutions (x1, x2, . . . , xn) does the above equation possess?
We shall see that there is a one-to-one correspondence between the nonnegative integral solutions of the equationx1+x2+. . .+ xn=k and the k- multisubset of n-set {a1, a2, . . . , an}.
If (x1, x2, . . . , xn) is a nonnegative integral solution of the equa- tion, we construct a k-multisubset of {a1, a2, . . . , an}by choosing the elementai with multiplicityxifor 1 ≤i≤n.Conversely, given ak-multisubset S of {a1, a2, . . . , an} we assign to the variable xi the value of the multiplicity of the element ai occurring in the k- multisubsetS.Hence the number of nonnegative integral solutions of the equation x1+x2+· · ·+xn=k is (n+k−1
k
).
Example 2.3.13:
Find the number ofpositive integral solutions of the equationx1+ x2+x3+x4 = 9.
We have a formula for the number ofnonnegative integral solu- tions of a linear homogeneous equation but here we have to deter- mine the number ofpositiveintegral solutions. To use our formula, we perform the following change of variables:
y1 =x1−1 y2 =x2−1 y3 =x3−1 y4 =x4−1.Plugging these into our initial equation, we get the equationy1+y2+y3+y4 = 5 where theyi variables are nonnegative. Note that (x1, x2, x3, x4) is a positive integral solution ofx1+x2+x3+x4 = 9 if and only if (y1, y2, y3, y4) is a nonnegative integral solution of the equation
2.3. PERMUTATIONS AND COMBINATIONS 75 y1 +y2 +y3 +y4 = 5. Hence the number of desired solutions is (4+5−1
5
)= 56.