Slide 1: PERMUTATION AND COMBINATION VISHWAS KHARE ASSISTANT PROFESSOR DEPT.OF MATHEMATICS S.S.R. COLLEGE SILVASSA UT OF DADRA NAGAR HAVELI email: vishwas_k2211@yahoo.co.in
Slide 2: FUNDAMENTAL PRINCIPAL OF COUNTING MULTIPLICATION PRINCIPLE If first operation can be done by m ways & second operation can be done by n ways Then total no of ways by which both operation can be done simultaneously =m x n ADDITION PRINCIPLE If a certain operation can be performed in m ways and another operation can be performed in n ways then the total number of ways in witch either of the two operation can be performed is m + n.
Slide 3: Example 1 Suppose Rakesh decide to go vapi and see movie with his friends . He can go vapi by 3 ways by car by auto or by bus and suppose 5 different movies are running in cinema hall. In how many ways he can go with his friends to vapi and see movie ? Ans first operation (to go vapi ) can be done by 3 ways (car, bus,auto ) Second operation (to see a movie) can be done by 5 ways (M1,M2,M3,M4,M5) Therefore he can go vapi with his friends and see movie by 3x5=15 different ways
Slide 4: EXAMPLE 2 How many 3 digit no can be formed by using digits 8,9,2,7 without repeating any digit? How many are greater than 800 ? A three digit number has three places to be filled Now hunderd’th place can be filled by 4 ways , After this tenth place can be filled by 3 ways After this unit place can be filled by 2 ways Total 3 digits no we can form =4x3x2= 24 Hundred place Tenth place Unit place
Slide 5: SECOND PART To find total number greater than 800 (by digits 8,9,2,7 ) (we observe that numbers like 827 , 972 etc. starting with either 8 or by 9 are greater than 800 in this case) Hence Hundred th place can be filled by 2 ways (by 8 or 9) After this tenth place can be filled by 3 ways After this unit place can be filled by 2 ways Total 3 digits no greater than 800 are =2x3x2=12 Hundred place Tenth place Unit place 8 9 2 7
Slide 6: EXAMPLE suppose you want to get a policy to get tax relief. Suppose 3 policy scheme available with L.I.C. and 5 policy schemes are available with Birla life insurance, in how many ways this can be done ?. ANS Using ADDITION PRINCIPLE we have 3+ 5=8 choices Note (1) that here first operation is to get policy from L.I.C. which can be done by 3 ways . And second operation means to get policy from Birla Life Insurance . (2) The meaning of words I operation and II operation changed according to the problem asked
Slide 7: There are 3 teams first consists of 5 members second team consist 4 members and third team consist 3 members . In how many ways they can sit for photograph so that members of same team are together . Now 3 teams can be placed by 3! ways (I st operation by m=3 ways) After teams are placed the members of team can interchange their positions by 5!x4!x3! Ways ( i.e. II operation by n= 5!x4!x3! Ways) Therefore Total number of ways = mxn = 3!x(5!x4!x3!) Team 1 Team 2 Team 3
Slide 8: PERMUTATION : A permutation of given objects is an arrangements of that objects in a specific order. Suppose we have three objects A,B,C. so there are 6 different permutations (or arrangements ) In PERMUTATATION order of objects is important . ABC ≠ ACB A C B A A A A A B B B B B C C C C C
Slide 9: PERMUTATION OF DISTINCT OBJECTS A permutation (or an arrangement) of n distinct objects taken r at a time is an ordered arrangement of r objects from these n objects ( 0 ≤ r ≤ n). the total number of different permutation of n distinct objects taken r at a time without repetition is denoted by n P r and given by n P r = Example Suppose we have 7 distinct objects and out of it we have to take 3 and arrange Then total number of possible arrangements would be 7 P 3 = = 840
Slide 10: Suppose there are n objects and we have to arrange all these objects taken all at the same time Then total number of such arrangements OR Total number of Permutation will be = n P n = = = n!
: EXAMPLE :- How many 3 digits number can be formed using digits 1,6,8,9,3,7 without repeating any digits ? Remark : 3 digits numbers may be 697,737,. . . . .etc .Here order of digits does matter ANS : we have total n=6 objects (digits ) and out of these 6 objects we have to select r=3 objects (digits) and arrange to form 3 digits number . Hence the number of such 3 digits number = 6 P 3 = = = 20
Slide 12: There posts chairman vice chairman and secretary are to be filled out of 10 suitable candidates . In how many different ways these posts can be filled . Remark :Anyone out of these 10 candidates become chairman similarly anyone of them can become vice chairman or secretary Solution : First chairman can be selected by 10 ways After this vice chairman can be selected by 9 ways ( because now only 9 candidates remains) After this secretary can be selected by 8 ways Therefore total number of different ways these posts can be filled is =10x9x8 =720
Slide 13: Q(1) In how many ways 2 Gents and 6 Ladies can sit in a row for a photograph if Gents are to occupy extreme positions ? SOLUTION Here 2 Gents can sit by =2! Ways ( As they can interchange there positions so first operation can be done by 2! Ways) After this 6 Ladies can sit by =6! Ways (Ladies can interchange their positions among themselves so second operation can be done by 6! Ways ) Hence total number of possible ways are = 2!x6! =1440 L L L L L L G G
Slide 14: In how many ways 3 boys and 5 girls sit in a row so that no two boys are together ? Girls can sit by 5! Ways After this now out of 6 possible places for boys to sit 3 boys can sit by 6 P 3 ways Hence total number of ways = 5!x 6 P 3 G G G G G
Slide 15: COMBINATION A combination is selection of objects in which order is immaterial Suppose out of 15 girls a team of 3 girls is to select for Rangoli competition Here it does not matter if a particular girl is selected in team in first selection or in second or in third . Here only it matter whether she is in team or not i . e. order of selection does not matter . In Permutation : Ordered Selection In combination : Selection ( Order does not matter)
Slide 16: SUPPOSE 3 OBJECTS A B C ARE THERE We have to select 2 objects to form a team Then possible selection ( or possible team ) AB ,AC,BC i.e. 3 different team can be formed Remark : Note that here team AB and BA is same OBJECTS A, B,C COMBINATIONS AB,BC,CA PERMUTATIONS AB,BA,BC,CB,AC,CA,
Slide 17: COMBINATION OF DISTINCT OBJECTS A combination of n distinct objects taken r at a time is a selection of r objects out of these n objects ( 0 ≤ r ≤ n). Then the total number of different combinations of n distinct objects taken r at a time without repetition is denoted by n C r and given by n C r = Suppose we have 7 distinct objects and out of it we have to select 3 . Then total number of possible arrangements would be 7 C 3 = = = = 35
Slide 18: EXAMPLE A Cricket team of eleven (11) players is to be formed from 20 players consisting of 7 bowlers , 3 wicket keepers and 10 batsmen. In how many ways the team can be formed so that it contains exactly 4 bowlers and 2 wicket keepers? Solution :- 4 bowlers can be selected out of 7 by = 7 C 4 ways 2 wicket keepers can be selected out of 3 by= 3 C 2 ways Remaining 6 batsman can be selected out of 10 by = 10 C 5 ways Hence total number of ways = 7 C 4 x 3 C 2 x 10 C 5
Slide 19: EXAMPLE In a box there are 7 pens and 5 pencils . If any 4 items are to be selected from these Find in how many ways we can select A) exactly 3 pens B) no pen C) at least one pen D) at most two pens Solution :- A) 7 C 3 x 5 C 1 B) 5 C 4 C) either 1 pen OR 2 pens OR 3 pens OR 4 pens 7 C 1 x 5 C 3 + 7 C 2 x 5 C 2 + 7 C 3 x 5 C 1 + 7 C 4 D ) either no pen OR 1 pens OR 2 pens 7 C 0 x 5 C 4 + 7 C 1 x 5 C 3 + 7 C 2 x 5 C 2