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PERMUTATION & COMBINATION

concept & examples on permutation & combination chapter.

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PERMUTATION & COMBINATION

  1. 1. PERMUTATION AND COMBINATION
  2. 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.
  3. 3. • 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
  4. 4. • 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
  5. 5. • 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
  6. 6. • 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
  7. 7. • 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 CB A A A A A B B B B B C CC C C
  8. 8. • PERMUTATION OF DISTINCT OBJECTS • The total number of different permutation of n distinct objects taken r at a time without repetition is denoted by nPr and given by • n Pr = where n!= 1x2x3x. . .xn • • 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 • 7P3 = = 840 • Where 7!= 7x6x5x4x3x2x1
  9. 9. • 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 Pn = = = n!
  10. 10. 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 = 6P3 = = 120
  11. 11. • Three 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
  12. 12. • 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 GG
  13. 13. • 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 6P3 ways • Hence total number of ways = 5!x 6P3 G G G G G
  14. 14. • 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)
  15. 15. 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,
  16. 16. • 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 Cr and given by • n Cr = • • Suppose we have 7 distinct objects and out of it we have to select 3 to form a team . • Then total number of possible selection would be • 7C3 = = = = 35 •
  17. 17. • 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 = 7C4 ways 2 wicket keepers can be selected out of 3 by= 3C2 ways Remaining 6 batsman can be selected out of 10 by = 10C5 ways Hence total number of ways = 7C4 x 3C2 x 10C5
  18. 18. • 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) 7C3 x5C1 • B) 5C4 • C) either 1 pen OR 2 pens OR 3 pens OR 4 pens • 7C1 x 5C3 + 7C2 x 5C2 + 7C3 x 5C1 + 7C4 • D) either no pen OR 1 pens OR 2 pens • 7C0 x 5C4 + 7C1 x 5C3 + 7C2 x 5C2

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concept & examples on permutation & combination chapter.

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