Permutations and-combinations-maths
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Transcript of Permutations and-combinations-maths
Xi ‘a’
V.MURUGAN
FUNDAMENTAL PRINCIPAL OF
COUNTING
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.
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 waysAfter this unit place can be filled by 2 waysTotal 3 digits no we can form =4x3x2= 24
Hundred place
Tenth place
Unit place
EXAMPLE :-
SECOND PARTTo 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
Both are ways to count the possibilitiesThe difference between them is whether
order matters or notConsider 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 permutationsIf no, then we are dealing with combinations
Permutations vs. Combinations
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 important . ABC ≠ ACB
A CB
A
A
A
A
A
B
B
B
B
B
C
CC
C
C
Permutations
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
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 ORTotal number of Permutation will be = n Pn
=
=
= n!NotationInstead of writing the whole formula,
people use different notations such as these:
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! = 1
Note: 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.
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
EXAMPLE :-
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
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 SelectionIn combination : Selection ( Order does not
matter)
COMBINATION
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 formedRemark : Note that here team AB and BA is same
OBJECTS A, B,C
COMBINATIONS
AB,BC,CA
PERMUTATIONS AB,BA,BC,CB,AC,
CA
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
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 selectA) 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
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
EXAMPLE:-