In our day to day life ‘Permutations and Combinations’ can help us to solve many problems. It is a very topic important for banking as well as various other competitive exams.
Permutations and Combinations have direct application on probability and various seating arrangements, distribution, selections. These questions are widely asked in almost every competitive exam such as banking, SSC, and Railways. Therefore, every competitive exam aspirant should focus on the basics of Permutations and Combinations and practice them. For all govt job notifications, check GovtJobPosts.in
Knowledge on factorials is crucial to start Permutations and Combinations,
Let’s start by revising the concepts of factorials and solve some problems on the same.
For a natural number, the product of all natural numbers up to the given number is called factorial and it is denoted by n!.
Thus, n! =n (n-1) (n-2)…3 x2x1 or it can also written as
n! =n (n-1)(n-2)!
n! =n (n-1)(n-2) (n-3)!
Factorials of some numbers
Note: There is no factorial for a negative number.
Applications of factorial:
Factorial is widely used in arrangement problems. It allows us to perceive however factorial helps in transcription things.
For example, we’ve five persons and we need to rearrange them on five vacant seats. We need to start with the primary place let’s say ‘A’. We select one person out of five for the primary place ‘A’. We can try this in five ways.
Now the other four seats are vacant and the four persons will arrange such that we are going to select one person out of four for the second place. We can try this arrangement in four ways.
In the same manner, for the third place, 3 ways, for the fourth place, two ways that and for the last vacant place only one means of choice is feasible. In general, for this is how we can arrange the people but if the number of people to arrange is large, it is very difficult to manually identify the possible ways to arrange the people. From the above example, we can say that five people can be arranged in 120 ways.
Mathematically it is represented as 5x4x3x2x1 which is 5! (5 factorial) = 120 ways
In short to arrange ‘m’ things at ‘m’ places; total arrangements made are equal to m!
Let’s try to solve the below examples.
Q.1) In how many ways can the letters of the word MUMBAI be rearranged?
Solution: It has a total of 6 words. So we will arrange 6 letters at 5 places in 6! = 720 ways.
But in this question, M is coming twice. If any letter is more than once in the word, then we divide the number of repetitions of the word. Therefore, we are dividing the total of 720 ways by 2! = 2.
So, totally different words that can be made will be 720/2 = 360.
Direct answer: 6!/2! = 360
Q.2) how many different words can be made using letters of LUCKNOW starting with L?
Solution: LUCKNOW has a total of 7 words. According to the question, L is fixed in the first place, so we will arrange the remaining 6 letters at 6 places in 6! = 720ways.
Permutations and Combinations:
Let’s start with some questions to understand the basic concepts of permutations and combinations
- When and how should we use Permutations and Combinations
- How we can use Permutations and Combinations to solve the problems?
By reading this article you should be able to understand the above questions and solve the related questions.
What is the difference between Permutations and Combinations?
Standing in a line, seated in a row
The distributed group is formed
Problems on digits
Problems on letters from a word
In short, each of the different arrangements that are made by taking few or all things is said to be a permutation.
Formula for calculating permutation is nPr = n!/(n-r)!
Suppose we have 5 children and we have to arrange them in 3 vacant places. Then, first of all, we will choose 3 children from 5. We can solve 5C3 different ways. After choosing 3 children, we will have to arrange them on the 3 vacant places, for that we will use a factorial concept. The total ways to arrange 3 children in 3 places are 3!
So total ways in which we can arrange 3 children from total 5 on 3 vacant places will be:
5C3x 3! = 5C2 x 3! = 5!/(3! x 2!) x 3! = 5!/2! = 60 ways.
Q.3) A boy & Girl captain are to be chosen out of an 11 student. In how many different ways we can choose them?
Solution: we have to select 2 captains from a total of 11 students. The ways of selection are 11C2 = (11 x 10)/(1 x 2) = 55. After selecting, we can arrange 2captain on 2 different positions in 2! = 2 ways. Such that total ways of selecting a captain will be = 11C2 x 2! = 55 x 2 = 110
Direct answer: 11P2 = 110
Q.4) In how many ways of the word EQUATION can be arranged so that all the vowels must come together?
Solution: In word ‘EQUATION’, we have 5 vowels (A, E, I, O, U) and 3 consonants (N, Q, T). In the given question, all five vowels should come together so that we can assume these 5 vowels to take one place together and the other 3 consonants will be put on 3 places, so total 4 places.
In order to arrange them in 4 places will be given by 4! = 24
We can arrange vowels in order as well as we can do that in 5! = 120 ways.
So total ways = 24 x 120 = 2880
Direct answer: 4!*5! = 24*120 = 2880
In short, each of the different groups or selections which can be made by some or all of a number of given objects without reference to the order of the objects in each group is said combination.
So,nCr = nCr = n! / [r! x (n-r)!]
nCr = nC(n-r)
Q.5) In a class there are 6 boys and 5 girls. In how many different ways we can choose the class monitor.
Solution: As we can clearly see that, we have to choose a student from a total of 11. So we will apply the combination concept to solve the problem.
11C1 = 11/1 = 11
Q.6) In a class, there are 6 boys and 5 girls. In how many different ways a boy and a girl can be selected for group leaders between two groups?
Solutions: We have to choose a boy from 6boys and a girl from 5 girls for two groups.
So total ways of selection = 6C1x5C1 = 6x5 = 30
Q.7) In how many different ways people can be selected from a total of 16 peoples?
Solution: We need to choose 11 peoples from a total of 16 peoples.
So the answer will be
16C11 = 16!/5!*(16-5)!
= 16!/5!*11! = (16*15*14*13*12)/(1*2*3*4*5) = 4368
Q.8) A bag contains 5 red and 3 blue marbles. In how many different ways, 2 red and 1 blue marble can be drawn?
Solution: The bag contains 5 red and we want to have 2 red marbles. So the number of ways of selecting red marbles =5C2 = 10
Similarly ways of selecting 1 blue marbles from 3 blue marbles = 3C1 = 3
So total ways to select 2 red and 1 blue marble will be = 10*3 = 30
Q.9) In how many different ways 11 (eleven) students can be selected from 15 students if 2 particular students are never selected?
Solution: It is given that 2 particular students are never selected. So we will do selection from the rest of the students which means we will select 11 students out of 13 students
So total ways of selection can be made
= (15-2)C11 = 13C11 it can also written as 13C2 = (12*13)/(2*1) = 78
Q.10) In how many different ways a team of 11 can be selected from 15 students if 2 particular students are always selected?
Solution: It is given that we have to select two particular students always, which means that we have a choice of selecting only for the remaining 9 students and the possible options are only 13.
So total ways of selection can be made
= (15-2)C(11-2) = 13C9 = 13C4 = (13 x12 x 11 x 10)/(1x 2 x 3 x 4) = 715
Key points related to Permutations and Combinations:
- If we have to arrange n things at n places, we can arrange them in n! Ways choices arrangement.
- Whenever we have to choose r things out of n, we have total nCr ways of selection among them.
- Whenever we have to choose r things from n and then also have to arrange those r things at r places, we have total nPr ways of selection.
Author: Ashwani Kumar
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