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# Permutations and-combinations-maths

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### Permutations and-combinations-maths

1. 1. 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.
2. 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 Unit Hundred Tenth place place place 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
3. 3.  SECOND PART To find total number greater than 800 (by digits 8,9,2,7 ) Hundred Tenth Unit place place place 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
4. 4.  Both are ways to count the possibilities The difference between them is whether order matters or not Consider a poker hand:  A , 5 , 7 , 10 , K Is that the same hand as:  K , 10 , 7 , 5 , A Does the order the cards are handed out matter?  If yes, then we are dealing with permutations  If no, then we are dealing with combinations
5. 5.  A permutation of given objects is an arrangements of that objects in a specific order. Suppose we have three objects A,B,C. A B C C A B A C B C B A B A C so there are 6 different permutations (or B C A arrangements ) In PERMUTATATION order of objects important . ABC ≠ ACB
6. 6.  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 nP = where n!= 1x2x3x. . .xn 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 7P3 = = 840 Where 7!= 7x6x5x4x3x2x1
7. 7.  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 = nP n = = = n! Notation Instead of writing the whole formula, people use different notations such as these:
8. 8. The factorial function (symbol: !) just means to multiply a series of descending natural numbers. Examples: 4! = 4 × 3 × 2 × 1 = 24 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 1! = 1Note: it is generally agreed that 0! = 1. It may seem funny that multiplying no numbers together gets you 1, but it helps simplify a lot of equations.
9. 9.  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 G L L L L L L G 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
10. 10.  In how many ways 3 boys and 5 girls sit in a row so that no two boys are together ? G G G G G 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
11. 11.  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)
12. 12. SUPPOSE 3 OBJECTS A B C ARE THEREWe have to select 2 objects to form a teamThen possible selection ( or possible team )AB ,AC,BCi.e. 3 different team can be formedRemark : Note that here team AB and BA is same OBJECTS A, B,C COMBINATIONS PERMUTATIONS AB,BC,CA AB,BA,BC,CB,AC,CA
13. 13.  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 nC r = 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
14. 14.  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 x 5C1 B) 5C4 C) either 1 pen OR 2 pens OR 3 pens OR 4 pens 7C 1 x 5C3 + 7C2 x 5C2 + 7C3 x 5C1 + 7C4 D) either no pen OR 1 pens OR 2 pens 7C x 5C4 + 7C1 x 5C3 + 7C2 x 5C2 0
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