Mathematical Methods for Physics and Engineering - Matematica.NET

PROBABILITYAnother useful result that may be derived using the binomial coefficients is thenumber of ways in which n distinguishable objects can be divided into m piles,with n i objects in the ith pile, i =1, 2,...,m (the ordering of objects within eachpile being unimportant). This may be straightforwardly calculated as follows. Wemay choose the n 1 objects in the first pile from the original n objects in n C n1 ways.The n 2 objects in the second pile can then be chosen from the n − n 1 remainingobjects in n−n1 C n2 ways, etc. We may continue in this fashion until we reach the(m − 1)th pile, which may be formed in n−n1−···−nm−2 C nm−1 ways. The remainingobjects then form the mth pile and so can only be ‘chosen’ in one way. Thus thetotal number of ways of dividing the original n objects into m piles is given bythe productN = n n−nC 1n1 C ···n−n1−···−nm−2 n2 C nm−1n! (n − n 1 )!=n 1 !(n − n 1 )! n 2 !(n − n 1 − n 2 )! ··· (n − n 1 − n 2 − ···− n m−2 )!n m−1 !(n − n 1 − n 2 − ···− n m−2 − n m−1 )!n! (n − n 1 )!=···(n − n 1 − n 2 − ···− n m−2 )!n 1 !(n − n 1 )! n 2 !(n − n 1 − n 2 )! n m−1 !n m !n!=n 1 !n 2 ! ···n m ! . (30.35)These numbers are called multinomial coefficients since (30.35) is the coefficient ofx n11 xn2 2 ···xnm m in the multinomial expansion of (x 1 + x 2 + ···+ x m ) n , i.e. for positiveinteger n∑(x 1 + x 2 + ···+ x m ) n n!=n 1 !n 2 ! ···n m ! xn1 1 xn2 2 ···xnm m .n 1 ,n 2 ,... ,nmn 1 +n 2 +···+nm=nFor the case m =2,n 1 = k, n 2 = n − k, (30.35) reduces to the binomial coefficientn C k . Furthermore, we note that the multinomial coefficient (30.35) is identical tothe expression (30.32) for the number of distinguishable permutations of n objects,n i of which are identical and of type i (for i =1, 2,...,mand n 1 +n 2 +···+n m = n).A few moments’ thought should convince the reader that the two expressions(30.35) and (30.32) must be identical.◮In the card game of bridge, each of four players is dealt 13 cards from a full pack of 52.What is the probability that each player is dealt an ace?From (30.35), the total number of distinct bridge dealings is 52!/(13!13!13!13!). However,the number of ways in which the four aces can be distributed with one in each hand is4!/(1!1!1!1!) = 4!; the remaining 48 cards can then be dealt out in 48!/(12!12!12!12!)ways. Thus the probability that each player receives an ace is4! 48! (13!) 4 24(13) 4=(12!) 4 52! (49)(50)(51)(52) =0.105. ◭As in the case of permutations we might ask how many combinations of kobjects can be chosen from n with replacement (repetition). To calculate this, we1136