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Permutation and Combination (joshi, Adhikari and Aryal)
We very often come across problems in which we have to compute the number of ways a set of objects can be arranged under some conditions. In this chapter, we discuss about the fundamental rules of counting along with permutation and combination. While arranging the objects we regard to order in some cases and we don't regard order in some cases. As examples concerning these problems , we can have , in particular in the probability theory. To tackle them we have to study different counting methods. The study of counting methods comes under the field of combinatorial algebra. Suppose there are 50 students in a class and we need to form a committee of 3 students . If a committee consists of the president , secretary, and treasurer we use permutation whereas if the post is not mentioned we use the combination. Similarly, three objects A, B, and C can be arranged in 6 different ways regarding the order. They are ABC, ACB, BCA, BAC , CAB, CBA. But if the order is not regarded all these arrangements have the same meaning. So the arrangement of objects regarding the order refers permutation and arrangements of objects without regarding order ( commonly known as selection) refers combination.
THE BASIC PRINCIPLES OF COUNTING ( Fundamental Rules of counting)
There are two fundamental rules of counting or selection based on the simple principles of multiplication and addition . Let one event can be performed in different m ways and another event can be performed in different n ways.If the task is independent i.e. completed in m ways independently then a total number of ways in which the task can be performed is m+n ways where as if the task is not completed by m ways or n ways independently then the total number of ways in which the task can be performed in m×n ways.
For instance, there are 6 busways and 4 railways from city P to Q . Then a person can travel from A to B in 6 ways by bus and 4 ways by train. Since travelling by bus is not followed by travelling by train , so the person can travel in 10 different ways from city P to city Q . In other words, the task is completed while travelling by bus or train. So, this is called addition principle of counting.
Likewise, suppose there are 4 gates to enter and 3 gates to exit In a football ground . Then a player can enter in 4 different ways and exit in 3 different ways . So the player can enter and exit from the ground in 4×3=12 different ways. In this case, the task is not completed by only one ; either entrance or exit.Hence, this is called multiplication principle of counting.
EXAMPLE;
A man has 3 shirts, 4 pants, and 2 caps . How many ways can he wear one of each at a time?
Solution:
Since there are 3 shirts , 4 pants
and 2 caps he can wear shirts in 3 different ways , pants in 4 different ways and caps in 2 different ways . So he can wear the dresses in 3×4×2=24 different ways.
FACTORIAL NOTATION
The product of the first n natural number is called factorial n or n factorial which is denoted by n!
So, n! = 1×2×3×…….. ×(n-1)×n
And n! = n (n-1)!
= n(n-1) (n-2)!
=n(n-1)………. 3×2×1
Value of 0!
Since n! += n(n-1)!
Putting n=1, we get
1! = 1×0!
`.` 0!= 1.
The factorial of negative integers , fractions, and irrational numbers is not defined.
PERMUTATION
If there are 'n' objects and they are to be placed in any definite arrangement or order ,then any given order of these 'n' objects is called permutation of 'n' objectives. Note that permutation is used if the position of the object within the group has to be taken into account.
Similarly, the arrangement of objects regarding the order is called permutation. While forming committee with specified post,while making numbers plates of the vehicles,telephone,recharge card number,in seat arrangements etc, we use permutation.
The various forms of permutation are;
P(n,n) = npn = \(\frac{n!} {(n-n)!}\)
P(n,r )= nPr = \(\frac{n!} {(n-r)!}\) (n is greater or equals to r)
P(n,r) = n^r
\(\frac {n!} {p!q!r!}\)
(n-1)!
Special case : Circular permutations of n objects to form necklace or bracelet or garland is
\(\frac{n-1} {2}\)
EXAMPLE
In how many different ways can one make a first , second ,third and fourth choice among 12 firms leasing construction equipment?
Here, n=12, r=4
Then the required permutation is:
12p4 =\(\frac {12!} {(12-4)!}\)= \(\frac {12!} {8!}\)= 12×11×10×9=11,880.
hence the required number of ways is 11,880.
COMBINATIONS
Combinations of objects mean just their collection without any regard to order or arrangement . The absence of order in the combination of objects makes it different from the permutations of the objects. There is only one combination of n objects: but for the same objects, the number of permutation is n!. Similarly, the number of combination of objects taken r at a time is less than the number of permutations of objects taken r at a time . The combination of n things taken r at a time is denoted by nr. Furthermore, the arrangement of objects without regarding the order is called combination. We use combinations while making committee, groups, while selecting questions , selecting balls etc. in which the order is not regarded . The arrangements ABC, ACB, BAC, CAB, CBA, BCA have the same meaning if the order is not regarded.
The number of ways in which r different objects can be chosen out of n different object is ;
C(n ,r) =nCr = \(\frac{n!} {r!{(n-r)!}}\)
Restricted Combinations
Example 1
A student has two libraries cards and there are 5 different books of mathematics and 3 different books of the economics of his interest . In how many ways can he issue one book of each subject ?
Solution:
Given,
Total number of mathematics books = 5
Total number of economics books = 3
For one book of each subject , he needs to select one book of mathematics and one book of economics . So he can make up his choice in ;
= \(\frac {5!}{4! 1!}\) ×\(\frac{3!}{1! 2!}\)
= 15 .
Therefore, he can make up his choice of books in 15 ways.
EXAMPLE 2
In an examination paper containing 12 questions a candidate has to answer 7 questions only : in how many ways can he choose the questions?
Solution:
The candidate can choose 7 questions out of 10 in C(10,7) wats, which gives
=\(\frac {10!}{{(10-7)}! 7!}\)
= \(\frac{8×9×10} {1×2×3}\)
=120.
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ASK ANY QUESTION ON Basic principle of counting
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