# MATHEMATICAL THEORY FOR SOCIAL SCIENTISTS THE BINOMIAL THEOREM. (p + q) 0 =1,

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1 THE BINOMIAL THEOREM Pascal s Triangle and the Binomial Expansion Consider the following binomial expansions: (p + q) 0 1, (p+q) 1 p+q, (p + q) p +pq + q, (p + q) 3 p 3 +3p q+3pq + q 3, (p + q) 4 p 4 +4p 3 q+6p q +4pq 3 + q 4, (p + q) 5 p 5 +5p 4 q+10p 3 q +10p q 3 +5pq 4 + q 5 The generic expansion is in the form of (p + q) n p n + np n 1 n(n 1) q + p n q n(n 1)(n ) + p n 3 q 3 + 3! n(n 1) (n r+1) + p n r q r + n(n 1)(n ) + p 3 q n 3 n(n 1) + p q n + npq n 1 + p n 3! In a tidier notation, this becomes (p + q) n n x0 (n x)!x! px q n x We can find the coefficient of the binomial expansions of successive degrees by the simple device known as Pascal s triangle: The numbers in each row but the first are obtained by adding two adjacent numbers in the row above The rule is true even for the units which border the triangle if we suppose that there are some invisible zeros extending indefinitely on either side of each row Instead of relying merely upon observation to establish the formula for the binomial expansion, we should prefer to derive the formula by algebraic methods Before 1

2 we do so, we must reaffirm some notions concerning permutations and combinations which are essential to a proper derivation Permutations Let us consider a set of three letters {a, b, c} and let us find the number of ways in which they can be can arranged in a distinct order We may pick any one of the three to put in the first position Either of the two remaining letters may be placed in the second position The third position must be filled by the unused letter With three ways of filling the first place, two of filling the second and only one way of filling the third, there are altogether different arrangements These arrangements or permutations are abc, acb, bac, cab, cba Now let us consider an unordered set of n objects denoted by {x i ; i 1,,n}, and let us ascertain how many different permutations arise in this case The answer can be found through a litany of questions and answers which we may denote by [(Q i,a i ); i 1,,n]: Q 1 : In how may ways can the first place be filled? A 1 : n Q : In how may ways can the second place be filled? A : n 1 ways, Q 3 : In how may ways can the third place be filled? A 3 : n ways, Q r : In how may ways can the rth place be filled? A r : n r ways, Q n : In how may ways can the nth place be filled? A n : 1 way If the concern is to distinguish all possible orderings, then we can recognise n(n 1)(n ) 31 different permutations of the objects We call this number n-factorial, which is written as, and it represents the number of permutations of n objects taken n at a time We also denote this by n P n We shall use the same question-and-answer approach in deriving several other important results concerning permutations First we shall ask Q: How many ordered sets can we recognise if r of the n objects are so alike as to be indistinguishable? A reasoned answer is as follows: A: Within any permutation there are r objects which are indistinguishable We can permute the r objects amongst themselves without noticing any differences Thus the suggestion that there might be permutations would overestimate the

3 number of distinguishable permutations by a factor of ; and, therefore, there are only recognizably distinct permutations Q: How many distinct permutations can we recognise if the n objects are divided into two sets of r and n r objects eg red billiard balls and white billiard balls where two objects in the same set are indistinguishable? A: By extending the previous argument, we should find that the answer is (n r)! Q: How many ways can we construct a permutation of r objects selected from a set of n objects? A: There are two ways of reaching the answer to this question The first is to use the litany of questions and answers which enabled us to discover the total number of permutations of n objects This time, however, we proceed no further than the question Q r, for the reason that there are no more than r places to fill The answer we seek is the number of way of filling the first r places The second way is to consider the number of distinct permutation of n objects when n r of them are indistinguishable We fail to distinguish amongst these objects because they all share the same characteristic which is that they have been omitted from the selection Either way, we conclude that the number is n P r n(n 1)(n ) (n r+1) (n r)! Combinations Combinations are selections of objects in which no attention is payed to the ordering The essential result is found in answer to just one question: Q: How many ways can we construct an unordered set of r objects selected from amongst n objects? A: Consider the total number of permutations of r objects selected from amongst n This is n P r But each of the permutations distinguishes the order of the r objects it comprises There are different orderings or permutations of r objects; and, to find the number of different selections or combinations when no attention is paid to the ordering, we must deflate the number n P r by a factor of Thus the total number of combinations is n C r 1 n P r n C n r 3 (n r)!

4 Notice that we have already derived precisely this number in answer to a seemingly different question concerning the number of recognizably distinct permutations of a set of n objects of which r were in one category and n r in another In the present case, there are also two such categories: the category of those objects which are included in the selection and the category of those which are excluded from it Example Given a set of n objects, we may define a so-called power set which is the set of all sets derived by making selections or combinations of these objects It is straightforward to deduce there are exactly n objects in the power set This is demonstrated by setting p q 1 in the binomial expansion above to reach the conclusion that n x ( ) n x x x!(n x)! Each element of this sum is the number of ways of selcting x objects from amongst n, and the sum is for all values of x 0,1,,n The Binomial Theorem Now we are in a position to derive our conclusion regarding the binomial theorem without, this time, having recourse to empirical induction Our object is to determine the coefficient associated with the generic term p x q n x in the expansion of (p + q) n (p+q)(p + q) (p+q), where the RHS displays the n factors which are to be multiplied together coefficients of the various elements of the expansion are as follows: p n The The coefficient is unity, since there is only one way of choosing n of the p s from amongst the n factors p n 1 q This term is the product of q selected from one of the factors and n 1 p s provided by the remaining factors There are n n C 1 ways of selecting the single q p n q p n r q r The coefficent associated with this term is the number of ways of selecting two q s from n factors which is n C n(n 1)/ The coefficent associated with this terms is the number of ways of selecting rq s from n factors which is n C r /{(n r)!} From such reasoning, it follows that (p + q) n p n + n C 1 p n 1 q + n C p n q + + n C r p n r q r + + n C n p q n + n C n 1 pq n 1 + q n n (n x)!x! px q n x x0 4

5 The Binomial Probability Distribution We wish to find, for example, the number of ways of getting a total of x heads in n tosses of a coin First we consider a single toss of the coin Let us take the ith toss, and let us denote the outcome by x i 1 if it is heads and by x i 0 if it is tails Heads might be described as a sucess, whence the probability of a sucess will be P (x i 1)p Tails might be described as a failure and the corresponding probability of this event is P (x i 0)1 p For a fair coin, we should have p 1 p 1,of course We are now able to define a probabilty function for the outcome of the ith trial This is f(x i )p x i (1 p) 1 x i with x i {0,1} The experiment of tossing a coin once is called a Bernoulli trial, in common with any other experiment with a random dichotomous outcome The corresponding probability function is called a point binomial If the coin is tossed n times, then the probability of any particular sequence of heads and tails, denoted by (x 1,x,,x n ) is given by P (x 1,x,,x n )f(x 1 )f(x ) f(x n ) p xi (1 p) n x i p x (1 p) n x, where we have defined x x i This result follows from the independence of thed Bernouilli trials whereby P (x i,x j )P(x i )P(x j ) is the probability of the occurrence of x i nad x j together in the sequence Altogether there are ( ) n x (n x)!x! nc x different sequences (x 1,x,,x n ) which contain x heads; so the probability of the event of x heads in n tosses is given by b(x; n, p) This is called the binomial probability function (n x)!x! px q n x 5

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