Overview(permutation and combination)
Before studying about permutation and combination, we will have to know about factorial.
Here we denote The product of the positive integers from 1 to n by n! And we read it as “factorial “
“The continued product of first n natural numbers, i.e., 1, 2, 3, …. (n – 1) n, we generally express this by the symbol n!
we read it as ‘factorial n”.” therefore, it is expressed as below
Hence, using the symbol of permutation, we get,
nPr = n (n – 1) (n – 2) … (n – r + 1)
Example: Get the value of the following factorials
1. 5!,
2. 6! − 5!
Sol.
(i) 5! = 5 × 4 × 3 × 2 × 1 = 120.
(ii) 6! − 5! = 6 × 5! − 5! = (6 − 1) × 5! = 5 × 120 = 600.
Fundamental principal of counting
The Sum Rule of permutation and combination
Let us consider two tasks which need to be completed.
If the first task can be completed in M different ways and the second in N different ways, and if these cannot be performed simultaneously, then there are M + N ways of doing either task. above all, This is the sum rule of counting.
The Product Rule of permutation and combination
Let’s suppose that a task comprises of two procedures. If the first procedure can be completed in M different ways and the second procedure can be done in N different ways after the first procedure is done, then the total number of ways of completing the task is M × N
Example
Suppose one woman or one man has to be selected for a game from a society comprising 17 men and 29 women.
In how many different ways can this selection be made?
Sol.
The first task of selecting a woman can be done in total 29 ways.
The second task of selecting a man can be done in 17 ways. It follows from the sum rule, that there are 17+29 = 46 ways of making this selection.
Permutations
“The different arrangements which can be made out of a given set of things, by taking some or all of them at a time are called permutations.”
Or
A permutation is an arrangement of objects in a definite order. Since we have already studied combinations, we can also interpret Permutations as ‘ordered combinations.
A permutation of n objects taken r at a time is an arrangement of r of the objects
(r≤n).
The symbols of permutations of n things taken r at a time are-
Note
1. The number of permutations of n different things taken r at a time in which p particular things never occur is-
2. The number of permutations of n different things taken r at a time in which p particular things are always present is-
Note- A permutation of n objects taken r at a time is also called r-permutation
Permutation
The arrangement of a set of n objects in a given order is called a permutation.
Any arrangement of any 
Therefore, The number of permutations of n objects taken r will be denoted as-
P(n, r)
Formula-
Permutation with repetitions
The number of permutations of n objects of which are q_1 are alike, q_2are alike, q_r
are alike is-
Sampling with or without replacement
Suppose we chose the samples with repetition, from example if we draw a ball from a urn then we put back that ball in the urn and again we pick a ball and we continue the process.
So this is the case of sampling with repetition then the product rule tells us that the number of such samples is-
n.n.n.n.n……….n.n =n^r
And if we pick a ball from the urn and we do not put it back to the urn, then this is the case of sampling without replacement.
Therefore, in this case the number of samples are-
Example
If i choose three cards one after the other from a 52-card deck. Find the number m of ways
We can do this in: (a) with replacement; (b) without replacement.
Sol.
(a) Each card can be chosen in 52 ways, so that, m = 52(52)(52) = 140 608.
(b) Here there is no replacement. he first card can be chosen in 52 ways, the second in 51 ways, and the third in 50 ways. So that
P(52, 3) = 52(51)(50) = 132 600
Example- There are 4 black, 3 green and 5 red balls. In how many ways we can arrange in a row?
Solution: Total number of balls = 4 black + 3 green + 5 red = 12
The black balls are alike,
The green balls are, and the red balls are alike,
Hence, the number of ways in which we can arrange the balls in a row =
Example
A box contains 10 light bulbs. Find the number n of ordered samples of:
(a) Size 3 with replacement,
(b) Size 3 without replacement.
Solution:
Example
In a sports broadcasting company, the manager must pick the top three goals of the month, from a list of ten goals. In how many ways we can select the top three goals?
Solution: Since the manager must decide the top three goals of the month, the order of the goals is very important! It decides the first-place winner, the runner-up, and the second runner-up. Thus, we can see that the problem is of permutation formula.
Picking up three goals from a list of ten:
Possible Permutations =
Therefore, there are 720 ways of picking the top three goals!
Circular permutations
A circular permutation of n objects is an arrangement of the objects around a circle.
If the n objects to be arrang round a circle we take an objects and fix it in one position.
Now the remaining (n – 1) objects can be arranged to fill the (n – 1) positions the circle in (n – 1)! ways.
Hence the number of circular permutations of n different objects = (n – 1)!
Example: In how many ways 8 girls can form a ring?
Sol.
If we keep one girl fixed in any position, remaining 7 girls, we can arrange them in 7! ways.
Hence, the required on. of ways
= 7 ! = 7. 6. 5. 4. 3. 2. 1 = 5040.
Combination
A combination of n objects taken at a time is an unordered selection of r of the n
Objects (r ≤n).
“The different groups or collection or selections that can be made of a given set of things by taking some or all of them at a time without any regard to the order of their arrangements are called their combinations.”
Here note that a combination of n objects taken r at a time is also called r-combination of n objects.
Therefore, The number of combinations are as-
Total number of combinations
Total number of combination of n different things taken 1, 2, 3 …. n at a time
Here note that-
Grouping
When we are require to form two groups out of (m + n) things, so that one group consists of m things and the other of n things. Now formation of one group represents the formation of the other group automatically.
Hence, the number of ways m things we can select from (m + n) things.
Example: If we need to choose for a competition from a class of 3 students, 4 students, in how many ways we can select them?
Sol.
The number of combinations will be-
therefore, these are the all ways that we can select them.
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