BLOQUE A. DIAGNÓSTICO
1.2. Territorios de violencia, espacios de la infancia amenazados
Suppose eight students (call them 1,2,3,...,8) are to be assigned to three projects (call them A, B, C); project A requires 4 students, project B requires 2, and project C also requires 2. In how many ways can the students be assigned to the projects?
To answer this, we could list all possible assignments like this:
A B C 1,2,3,4 5,6 7,8 1,2,3,4 7,8 5,6 1,2,3,4 5,7 6,8 1,2,3,4 6,8 5,7 1,2,3,5 4,6 7,8 . . . . . .
However, the deadline for the projects will probably have passed by the time we have finished writing down the complete list (in fact there are 420 possible assignments). We need a nice way of counting such things.
Each assignment is what is called an ordered partition of the set {1,2,...,8} into subsets A,B,C of sizes 4, 2, 2. Here is the general definition of such a thing.
DEFINITION Let n be a positive integer, and let S = {1,2,...,n}. Apartition of S is a collection of subsets S1, . . . ,Sksuch that each element of S lies in exactly one of these subsets. The partition is ordered if we take account of the order in which the subsets are written.
The point about the order is that, for instance in the above example, the ordered partition
{1,2,3,4} {5,6} {7,8} is different from the ordered partition
{1,2,3,4} {7,8} {5,6}
even though the subsets involved are the same in both cases.
If r1, . . . ,rkare non-negative integers such that n = r1+∙∙∙ + rk, we denote the total number of ordered partitions of S = {1,2,...,n} into subsets S1, . . . ,Sk of sizes r1, . . . ,rkby the symbol
n
r1, . . . ,rk
COUNTING AND CHOOSING 135 Example 16.4
(1) The number of possible project assignments in the example above is
8 4,2,2
.
(2) If Alfred, Barney, Cedric and Dugald play bridge, the total number of different possible hands that can be dealt is
52 13,13,13,13
. (3) It is rather clear that
n r,n−r =n r . PROPOSITION 16.4 We have n r1, . . . ,rk = n! r1!r2!...rk! .
PROOF We count then! arrangements of S = {1,2,...,n} in stages as follows:
Stage 0: Choose an ordered partition of S into subsets S1, . . . ,Skof sizes
r1, . . . ,rk; the number of ways of doing this is n
r1, . . . ,rk
.
Stage 1: Choose an arrangement of S1: there arer1! choices.
Stage 2: Choose an arrangement of S2: there arer2! choices. And so on, until
Stage k: Choose an arrangement of Sk: there arerk! choices. By the Multiplication Principle, we conclude that
n! =r1, . . . ,n r
k
r1!...rk! The result follows.
136 A CONCISE INTRODUCTION TO PURE MATHEMATICS
Example 16.5
(1) The number of project assignments in the first example is 8
4,2,2
= 8!
4!2!2! =420 . (2) The total number of bridge hands, namely
52 13,13,13,13
,
is approximately5.365×1028; quite a large number. The numbers n
r1,...,rk are called multinomial coefficients, for the following
reason.
THEOREM 16.3 Multinomial Theorem
Let n be a positive integer, and let x1, . . . ,xk be real numbers. Then the expansion of (x1+∙∙∙ + xk)n is the sum of all terms of the form
n r1, . . . ,rk xr1 1 . . .xrkk
wherer1, . . . ,rk are non-negative integers such that r1+∙∙∙ + rk=n. PROOF Consider
(x1+∙∙∙ + xk)n= (x1+∙∙∙ + xk) (x1+∙∙∙ + xk) . . . (x1+∙∙∙ + xk) . In expanding this, we get a termxr11. . .xkrk by choosing x1fromr1of the brackets,x2from r2brackets, and so on. The number of ways of doing
this is
n r1, . . . ,rk
,
so this is the coefficient ofxr11. . .xkrk in the expansion. Example 16.6
(1) The expansion of (x+y+z)3 is
(x+y+z)3=x3+y3+z3+3x2y+3xy2+3x2z+3xz2+3y2z+3yz2+6xyz . (2) The coefficient ofx2y3z2in the expansion of (x+y+z)7is
7 2,3,2
= 7!
COUNTING AND CHOOSING 137 (3) Find the coefficient ofx3in the expansion of (1−x13+2x2)5.
Answer A typical term in this expansion is
5
a,b,c
∙ 1a∙ (−1x3)b∙ (2x2)c
wherea+b+c = 5 (and a,b,c ≥ 0). To make this a term in x3, we need
−3b+2c = 3 and a+b+c = 5 .
From the first equation, 3 dividesc, so c = 0 or 3. If c = 0 then b = −1, which is impossible. Hence c = 3, and it follows that a = 1,b = 1. Thus there is just one term inx3, namely
5 1,1,3
(−1
x3)(2x2)3=−160x3.
In other words, the coefficient is−160.
Exercises for Chapter 16
1. Evaluate the binomial coefficients 83 and 155.
2. Liebeck, Einstein and Hawking pinch their jokes from a joke book which contains 12 jokes. Each year Liebeck tells six jokes, Einstein tells four and Hawking tells two (and everyone tells different jokes). For how many years can they go on, never telling the same three sets of jokes? 3. Josephine lives in the lovely city of Blockville. Every day Josephine
walks from her home to Blockville High School, which is located 10 blocks east and 14 blocks north from home. She always takes a shortest walk of 24 blocks.
(a) How many different walks are possible?
(b) 4 blocks east and 5 blocks north of Josephine’s home lives Jemima, her best friend. How many different walks to school are possible for Josephine if she meets Jemima at Jemima’s home on the way?
(c) There is a park 3 blocks east and 6 blocks north of Jemima’s home. How many walks to school are possible for Josephine if she meets Jemima at Jemima’s home and they then stop in the park on the way?
4. (a) Prove that
n+1 r =n r + n r −1 .
138 A CONCISE INTRODUCTION TO PURE MATHEMATICS
(b) Prove that for any positive integer n, 3n=
∑
n k=0 n k 2k.5. Three tickets are chosen from a set of 100 tickets numbered 1,2,3,..., 100. Find the number of choices such that the numbers on the three tickets are
(a) in arithmetic progression (i.e., a,a+d,a+2d for some a,d) (b) in geometric progression (i.e., a,ar,ar2for some a,r).
6. The digits 1,2,3,4,5,6 are written down in some order to form a six- digit number.
(a) How many such six-digit numbers are there altogether? (b) How many such numbers are even?
(c) How many are divisible by 4?
(d) How many are divisible by 8? (Hint: First show that the remainder on dividing a six-digit number abcde f by 8 is 4d +2e+ f .)
7. (a) Find the coefficient of x15in (1+x)18.
(b) Find the coefficient of x4in (2x3− 1
x2)8.
(c) Find the constant term in the expansion of (y+x2− 1
xy)10.
8. The rules of a lottery are as follows: You select 10 numbers between 1 and 50. On lottery night, celebrity mathematician Richard Thomas chooses at random 6 “correct” numbers. If your 10 numbers include all 6 correct ones, you win.
Work out your chance of winning the lottery.
9. Here’s another way to prove Fermat’s Little Theorem. Let p be a prime number.
(a) Show that if r,s are positive integers such that s divides r, p divides r and p does not divide s, then p divides r
s.
(b) Deduce that p divides the binomial coefficient p
k for any k such that 1 ≤ k ≤ p−1.
(c) Now use the Binomial Theorem to prove by induction on n that p divides np− n for all positiveintegers n. Hence, deduce Fermat’s Little Theorem.
COUNTING AND CHOOSING 139 10. n points are placed on a circle, and each pair of points is joined by a straight line. The points are chosen so that no three of these lines pass through the same point. Let rnbe the number of regions into which the interior of the circle is divided.
Draw pictures to calculate rnfor some small values of n. Conjecture a formula for rnin terms of n.
11. Prove that if rnis as in the previous question, then for any n,
rn=1+n 2 +n 4 . Was your conjecture in the previous question correct?
12. The other day, critic Ivor Smallbrain gave a lecture to an audience con- sisting of five mathematicians. Each mathematician fell asleep exactly twice during the lecture. For each pair of mathematicians, there was a moment during the lecture when they were both asleep. Prove that there was a moment when three of the mathematicians were simultaneously asleep.