Introduction:
Permutation: The ways of arranging
or selecting smaller or equal number of persons or objects from a group of
persons or objects with regard to the order of arrangement or selection, are
called permutations.
In simple words, it means
arrangement of things where the order of things is considered.
Let us consider an Example,
We have to form a
number consisting of three digits using the digits 1, 2, 3. The total number of
Possible arrangement will be =6 (Factorial 3 ie.3x2x1):
Alternative 1: (1, 2, 3)
Alternative 2: (2, 3, 1)
Alternative 3: (3, 1, 2)
Alternative 4: (1, 3, 2)
Alternative 5: (2, 1, 3)
Alternative 6: (3, 2, 1)
Each one of these
possibilities is called permutation of three digits taken all at a time.
Permutations can also be calculated by using the following
formula:
^{n}P_{r}
= n (n1) (n2)……. (nr+1)
or ^{n}P_{r}
= n! / (nr)!
Where n = total no. of things
r =things to be chosen
Again, if we have to form a number
consisting of two digits using the digits 1, 2, 3. The total number of Possible
arrangement following the above formula will be =6
Alternative 1: (1, 2)
Alternative 2: (2, 3)
Alternative 3: (3, 1)
Alternative 4: (1, 3)
Alternative 5: (2, 1)
Alternative 6: (3, 2)
Each one of these
possibilities is called permutation of two digits taken at a time.
Fundamental Principles of counting:
(a) Multiplication Rule: If a certain thing can be
done in ‘m’ different ways and when it has been done, a second thing can be
done in ‘n’ different ways then total number of ways of doing both things
simultaneously be = m x n.
(And) = Multiply
E.g.
If one can go to school by 5 different buses and then come back by 4 different
buses then total number of ways of going to and coming back from school = 5x4 =
20 ways.
(b) Addition Rule: If there are two different jobs
which can be done in ‘m’ ways and in ‘n’ ways respectively then either of two
jobs can be done in (m + n) ways.
(Or) =
Add
E.g.
If one wants to go school either by 5 buses or to by auto where there are 4
autos, then total number of ways of going = 5+4 = 20 ways.
Practical Approach:
A) Permutation of things when they all are
different: To find the total number of permutation of n different things
taken n or r at a time will be:
^{n}P_{r} = n! /
(nr)!
Ex.1. In how many ways the word "Square” can be arranged taking
all at a time.
Ans: Since there is 6 different letters, the number of
permutations of choosing all at a time equals
^{6}P_{6}=6x5x4x3x2x1
=720 ways
Ex.2. In how many ways the word "Square” can be arranged taking 3
at a time.
Ans: Since there is 6 different letters, the number of
permutations of choosing 3 at a time equals
^{6}P_{3}=6x5x4
=120 ways
B) Permutation of things when they all are not
different: To find the total number of permutation of n different things
taken all at a time will be:
^{n}P_{r} = n! /
(p)! (q)! (r)!
Ex.1. In how many ways the word "Jalpaiguri” can be arranged
taking all at a time.
Ans: Since there is 10 letters where A and I is repeated 2 times,
the number of permutations of choosing all at a time equals = 10! / (2)! 2)! = 9, 07,200 ways.
C) Permutation of things which may be repeated:
To find the total number of permutation of n different things in which any item
can be repeated without restriction, the total number of possible arrangement
will be ^{n}P_{r} = n^{3}
The
above point will be clear by the following example.
Ex.1. In a quiz competition there are 6 students. In how many ways
1^{st}, 2^{nd} and 3^{rd} prizes can be awarded to 6
students.
Again if there are three prizes, one in quiz, one in sport and one
in drawing. In how many ways these prizes can be distributed.
Ans: 1^{st} case: Since prizes are based on position,
1^{st}
prize can be awarded in 6 ways
2^{nd}
prize can be awarded in 5 ways
3^{rd}
prize can be awarded in 4 ways.
Total no of
permutations will be = 120 ways (^{n}P_{r})
2^{nd}
case: Since prizes are based on competition, all the prizes can be awarded to
same student i.e.
1^{st}
prize can be awarded in 6 ways to 6 students.
2^{nd}
prize can be awarded in 6 ways to 6 students.
3^{rd}
prize can be awarded in 6 ways to 6 students.
Total no of
permutations will be: n^{3}= 6^{3}=216 ways
D) Permutation in a ring or in a circle: To
find the total number of permutation of n different things in which any item
can be repeated without restriction, the total number of possible arrangement
will be = (n1)!
There are three different cases. Consider the following example:
Ex.1. In how many ways can 6 persons be arranged at a round table
so that 2 particular persons may sit together?
Ans: Arrangement with respect to table: Taking 2 persons as
one person, we are to arrange 5 (=1+4) persons in 5! ways. Again 2 persons may
sit amongst themselves in 2! Ways.
Therefore, the reqd. no. of arrangement = 5! X 2! = 240 ways
Arrangement with respect to each other: at first 2 particular persons can
be arranged themselves in 2! Ways. Keeping them fixed and taking as one person,
all the 5 persons (including remaining 4 persons) can be arranged in (51)=4!
Ways.
Therefore, the reqd. no. of arrangement = 4! X 2! = 48 ways
Again, Suppose that we have to arrange 6 persons in such a way
that no person has the same neighbours in both clockwise and anticlock wise
direction, in such a case the required no. of permutations will be = ½*(61)! =
60 ways.
E) Restricted Permutation:
a) To find the total number of permutation of n
different things taken r at a time in which p particular things never occur is:
^{np }P _{r}.
Ex.1.
In how many ways 10 things can be arranged taking 3 at a time in which 2
particular things never occur?
Ans: n= 10, r=3 and p=2,
Therefore, number of ways = ^{np
}P _{r }=^{ 102 }P_{3}=^{8 }P_{3}=336
ways
b) To find the total number of permutation of n different things
taken r at a time in which p particular things occupy stated place is: ^{n –
p }P _{r  p}.
Ex.1. Find the number of different numbers
of 3 digits can be formed with the digits 1, 2,3,4,5 in which the unit place is
always occupied by 5.
Ans: n= 5, r=3 and p=1,
Therefore, number of ways = ^{n –
p }P_{r  p} =^{ 51 }P_{31}=^{4 }P_{2}=12
ways
c) To find the total number of permutation of n different things
taken r at a time in which p particular things always present is: ^{(n – p)}P_{(r
– p)} x ^{ r }P_{p}.
Ex.1.
In how many ways 10 things can be arranged taking 3 at a time in which 2
particular things always occur?
Ans: n=10, r=3 and p=2,
Therefore, number of ways = ^{np
}P _{r –p} x ^{r }P_{p}^{ }=^{ 102 }P
_{3 –2} x ^{3 }P_{2}^{ }
^{
}=^{8
}P _{1} x ^{3 }P_{2}^{ }=8 x 3 =24^{ }
d) To find the total number of permutation of n different things
taken r at a time in which p particular things always occur together in an
assigned order is: (rp+1) x ^{(n – p)}P_{(r – p)} .
Ex.1. Find the number of different
numbers of 3 digits can be formed with the digits 1, 2,3,4,5 in which (12) is
always together and also in the order given.
Ans: n= 5, r=3 and p=2,
Therefore, number of ways = ^{(}rp+1)
x ^{(n – p)} P_{(r – p)} = (32+1) x ^{(5 – 2)} P_{
(3 – 2)}
= (4) x ^{(3)} P_{ (1)}
= 4 x 3 = 12 ways
Note:
if 12 are not in the order, then the number of ways will be
=12 x 2! = 24 ways
Here
2 are for arrangement of 1and 2 in two ways.
e) To find the total number of permutation of n different things
taken all at a time in which p particular things will occur in assigned order
is: ^{n}P_{r }= n ! / P!
Ex.1.
If the word NOT be arranged with three letters in such a way that O should be
placed before T, then what will be the total number of permutations.
=3! / 2! = 3 ways.
(Permutations
are NOT, OTN, ONT.)
