‘The standard deviation S= (C) (150)? 1430 = oon. 199 2 (10) = 10(2.573) * S=25.73 i COMBINATORIAL ANALYSIS Combinatorial Analysis is the mathematical theory of counting. Many problems in probability theory can be solved by counting the number of different ways that a certain event can occur. 193o1g | UNIVERSITY ELEMENETARY MAJ u MATICS 2 y ‘Combinatorial analysis includes the study of permutations, combinations and partitions. It is also concerned with determining the number of logical possibilities of some event without necessarily enumerating each case. The basic principle of counting Theorem: Suppose an operation consists of two steps, of which the first can be made in n; ways and for each of these the second can be made in np ways, then there are nj.n) ways of making the whole operation. Proof: The basic principle may be proved by first defining the ordered pair (m;, y;) to be the outcome which arises when the first step results in possibility yi. The set of all possible outcomes is therefore composed of the following nj.n2 pairs. (x1, 1), Ot Ya)yeseeeeeee 3015 Yng) (x2, ¥1), 2; Vay ‘2s Yng) (Xn Yas ny Vadyereeeeee s(%ry> ng)2018 | UNIVERSITY. ELEMENETARY MATHEMATICS 2 Solution Let B, and Bz represent the two blouses and s}, 82, 3 and sq represent the four skirts. blouse and then a skirt are shown below: The various ways of choosing a S; (Bi, Si) Bi 2 (Bi, S) 3 (Bi, Ss) Sa (Bi, Sa) 1 (Bz, Si) - 2 (Bz, S2) 3 (Bz, S3) S4 (Bz, Sa) Definition For any positive integer n, n! is read “n factorial” and is defined by n! =n(n—- 1)(n—-2) .........4 (2)Q) Also 0! =1 196» 2018 | UNIVERSITY ELEMENETARY MATHEMATICS 2 permutations permutation is any arrangement of r objects selected from a single group of n possible objects. The arrangement of any rs 1 of these objects in a given order is called a permutation of the n objects taken r at a time. In permutation, both composition and order are important, The number of permutations of n objects taken r at a time is denoted by »P, ni ~ (n-r)! Where nis the total number of objects ris the number of objects selected Also, “P,=n(n-1). "P, =n(n— 1) (n—2)....... l=n! This implies that there are n! Permutations of n objects, taken (n—r + 1) and when r= n, then all at a time. . Permutations with Repetitions The number of permutations of n objects of which nj are alike, nz are alike,...... 5 Ny is n! ny IMg!...M, 197 og italEMATICS 2 ERSITY ELEMENETARY M 2018 | ole ~ aqtment in L. AUTECH consists of 10 lecturers, each of , were attached 3 project students. Sf one lecturer and one xt student are to be chosen as best lecturer and student of ’ "ear, how many different choices are possible? $ stioD: see from the basic principle of counting that fhere are 10x a0) possible choices gree Diagram ‘The t7ee diagram is a graph that is helpful in organizing calculations that involve severah stages. Each segment in the tree is one stage of the problem. ‘The branches of a tree diagram are weighted by probabilities. The tree is constructed from left or right, and the number of branches at each point corresponds to the number of ways the next event can occur. Example: Ifa lady has two blouses and four skirts. In how many ways can she choose a blouse and a skirt? 195MA EMENETARY 2018) UNL Exampl (1) Given just three letters A, B and C, how pr permufations of size 3 can we get? Solution: The number of permutations of letters A, B, C taken three at a 3 "3 time is: 3! 1.23 Ps Gayot The list of permutations of all three elements is: ABC, BCA, ACB, CAB, BAC, CBA (2) Given the first five letters of the alphabet, determine the number of permutations of the five elements taken three at a time. 5 Pa, Solution The number of permutations of the five alphabets taken three at a time: 5! 5P3 = = 60 (3) The hon Machine shop Ltd. had eight screw machines but only three spaces available in the production area for the machines. In how many different ways can the eight machines be arranged in the three spaces available? BP, 1982018| UNIVERSITY ARY MATHEMATICS 2 n= 8 machines and r= 3 Spaces available, then _. 8.7.6,51 gg = 336 ic. there are 336 different Possible arrangements, (4) Bayo has 12 books that he wants to arrange on a shelf. 6 are Of these, 4 are different Mathematics books, different Physics books while 2 are different Chemistry books. How many different arrangements are possible if (i) The books in each particular subject must 4Il stand together? 2 (ii) Only the Mathematics books must stand together? - Solution: (i) The Mathematics books can be arranged among themselves in ‘py = 4! Ways. The Physics books in °P, = 6! Ways The Chemistry books in *P,= 2! Ways The three groups of books can be arranged in 5P; = 3! Ways Therefore, the required number of arrangements = 4! 6! 2! 3! = 207,360 1992018 | UNE sider the 4 Mathemati k which can be arranged in °, 2 i) Con ics books as one big book. (ii Then we have 4 bool 9! Ways. In all of these ways, only the Mathematics books are together. But the Mathematics books can be arranged among themselves in “P, = 4! Ways. Then the required number of arrangements = 9! 4! = 8709120 (5) A chess tournament has 10° competitors of which 4 are Hausas, 3 are Ibos, 2 are Yoruba and 1 is Efik. If the tournament result lists just the ethnic tribe of the players in the order in which they played. How many outcomes are possible? Solution: = 12,600 possible outcomes 41312! (6) How many distinct permutations can be found from all the letters of the word: sail There are: Solution: For “Sociological” the number of permutations of the letters in the word is 12! 3! 212! 211! 114! Be 2002018 | UNIVERSITY ELEMENETARY MATHEMATICS 2 Since there are 12 letters of which 3 are 0, 2 are ©, 2 are i and 2are 1 Combinations A combination of n different objects taken r at a time is a selection of r out of the n objects with no attention given to the order of arrangement. Combination is concemed with determining the n number of different groups of r objects that could be formed from a total of n objects. The number of combinations of n objects taken r at a time is —1)...(n=T +1) denoted by "C, or (@) and is given by "C,= eee rt nl : ri(n-r)! forr¢n "C, or @) represents the number of different groups of size r that could be selected from a set of n objects when the order of selection is not considered relevant. Examples: (1) Evaluate (a) "C3 (b) “Cx Solution: cH Tale! Tl _ 7654824 35 @) C= S@-a “Wa” Saaaa2d tcej= —t 2 4 = Isince ol =1 (0) "C= Fama ai a 2012018 | UNIV 2) How many committees of 3 can be formed fi ( people? taken 3 at a time. 84746 4 oe = Thus: °C: * 33304 56 ie. 56 different committees can be formed. (3) From a group of 5 girls and 7 boys, how many dif committees consisting of 2 girls and 3 boys formed, What if2 of the boys are feeding and re serve on the committee together? Solution: 48 The 2 girls can be chosen from the 5 girls in ways boys can be chosen in ways. Hence, the committee ¢ chosen in sata = 350 possible committees consisting of 2 girls and 3 If 2 of the boys refuse to serve on the committee there are (7C,)(°C3) possible groups of 3 boys n 202 aversity ELEME Ary MATHEMATICS 2 ry ELEMEN! a VERS >) groups of 3 boys 2018 | UNI an. jther of the 2 fending boys and (Cy) ending boys ‘containing exactly | of the fending cr follows that there are: COEC?) + CC)NCC2 . 6 . of 3 boys not containing both of the fending boys. Since there are 5€ ways to choose 2 girls, it follows that in in this case, there are 30(°C2) = 300 possible committees. Exercises (1) Evaluate the followings (i) "Ps: GFP, Gil) °C; (iv) *Ce (v)"P3 (vi) *Cs @) In how many ways can 3 Indians, 4 Nigerians, 4 Ghanaian and 2 Americans be seated in a row so that those of the same nationality sit together? (3) A farmer buys 3 goats, 2 pigs and:4 rabbits from a man who has 8 goats, 4 pigs and 10 rabbits. How many choices does the farmer have? (4) A committees of 4 is to be formed from a group of 20 people. How many different committees are possible? (5) In how 4 ‘ow many ways can 7 people be Seated at a round table ie (a) (b) They can sit anywhere? 2 art p . Particular people Must not sit next to each other?