Permutation & Combination – 12 Problem types that can come in IIT JEE - JEE Main 2025

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Permutation & Combination is a very important but slightly confusing topic in Mathematics. A lot of students find it difficult to really understand the difference between Selection (Combination) and arrangement (Permutation) in this chapter. Making this video of Permutation & Combination took tremendous amount of effort to build up as well. In order to make this chapter super simple, we have started this video with weightage of Permutation & Combination in JEE and then we systematically go into the syllabus comparison between JEE Main 2024 and JEE Main 2025. This will ensure you have an exact idea of the differences.

Then we will jump into the Mind map of the chapter, so you know which topics and subtopics are covered in Permutation & Combination chapter for IIT JEE - JEE Main and JEE Advanced, and how these are interlinked with each other. That helps you cover the entire Permutation & Combination chapter comprehensively and ensure you do not miss any single subtopic or concept from it.

Then we'll go into 12 problem type categories in which I have classified all the last 5 to 10 years of questions that came in JEE Main. Since the video has 5-6 questions from each of the 12 types of problems in Permutation & Combination, mostly from Past Year Questions of JEE Main, we are keeping 2 questions per type in the blog. We will add the list of the 12 problem types followed by 2 questions from each type.

For every topic we will go into the subtopics that need to be covered then go into more deeper points of what exactly there is in the syllabus using Mind maps and ultimately come on to the 12 problem types that can be asked in Permutation & Combination chapter.

Before we start this video there is a “do” and there is a “don't”. The “do” is do not watch this video like you are watching a movie but keep making notes, keep solving questions, pause the video and see how you can arrive on an answer and develop your approach while solving. The “don't” is that this video is not meant for a fresher student who has not just done Permutation & Combination. We will not go into very Basics but yes, I will tell you the concept wherever and whenever it is required.

Permutation & Combination is one such Topic in mathematics which is usually a hit or miss for students. There are few students who love this topic. They are able to solve questions generally because their raw thinking ability is very high while there are other students who are not very comfortable with this topic. Whatever kind of classes they do or even questions they practice, still when the question is framed or put in the exam, they find it difficult to handle it. They are not able to figure out which formulae to apply where.

So, if you are one of such students who find this topic difficult and unable to do Permutation & Combination questions, then this video is going to simplify your life to a great extent. As a faculty with over 11 years of experience in JEE Mathematics, I have come up with a unique teaching approach in this topic that is based on 24 variations.

Questions in P&C can be written in 24 different variations. Once you understand how the same question, which has 24 different variations can be handled in each of its avatar, you understand what technique to apply there and you develop a lot of confidence in P&C. Then if you practice questions based on PYQs largely, you will be able to handle questions based on P&C easily.

So, this video starts with the idea of syllabus comparation and then we move towards those 24 variations which are something unique and I have developed based on my own experience. We understand all the problem-solving techniques in this chapter and then we move towards the survey of problems of JEE Mains for the last 3-4 years. In that I have divided the problems into 12 problem categories. In each of these problem categories you will understand what is the type of problem and what are the techniques that we can use in that problem category to come up with an answer.

So let us start with a Syllabus comparison of 2023 and 2024 to 2025. If you look into this syllabus, you can see virtually there is no change that has been done. The important topic remains Fundamental principle of counting, Permutation as Arrangement, Combination as Selection, use of nPr and nCr, then simple applications.

This does not exactly paint what constitutes this chapter or what kind of problems and what kind of exact theories are in this chapter. So let us go into it using a Mindmap.

Mindmap of the chapter – Permutation & Combination

If I look into this Mindmap of P&C, you will notice that the discussion starts with fundamental principle of counting, where we understand when to multiply things, when to add things, what is the box method and how to apply box method to solve questions based on numbers. And you will ultimately see that these number-based questions are some of the most important questions asked in this chapter. They have been asked so many times. Many types of questions are there whom we'll discuss it in detail.

Next, we move on to the idea of Permutation where we understand the formula nPr and what are the validity or important parameters when the formula is going to hold. Most students know what is nPr but they do not know what is the validity of this formula. If repetition happens in objects, how can this formula not be applied and then what techniques to use becomes another discussion in permutation.

Then we move towards Combination. Then there are restricted P&C based questions. Restricted permutation is when you place some condition on Arrangement, for example, this should be the first letter of the word or this should be the last letter of the word or these three letters should come together or these letters should never come together.

All these kinds of questions we handle using string method & Gap method. Then there are questions based on words like rank-based questions that we need to understand how to solve them. Then in restricted combination like we had restricted arrangement, we have restricted selection. Again, some conditions will be placed like what kind of selections cannot be made and how to make those Selections in such kind of questions. Again, you have questions which are based on at least or at most constraint. So, in such kind of questions, you need to make cases.

We will see a category of problems, how those questions are being made by JEE and how to answer those questions. Then there are questions based on number of divisors and sets and

ultimately total number of selections. Then we move towards the idea of distribution and here you can see there are two types of primary distribution - one in which the objects are distinct and one in which the objects are non-distinct. So, in the first case there is a particular method of solving, and in the second case there is Beggar's method which we generally use to solve questions. Then our discussion moves towards Circular permutation and de-arrangement.

These are all the general kind of questions and theories in this chapter but to give you a more tangible idea, we are going to take something very unique that I have developed. How can we take one problem and write 24 variations.

The major issue in this chapter that most of the students will face is despite knowing the basic theory and basic problem types, they get confused how to solve and which method to use where. Also to notice what kind of changes are coming in the question and how to see through those variations. So, I have come up with an approach known as 24 variations. Using this you will understand how can we take one question, write 24 variations to that question, and what should be the technique that will be valid in each of those variations. This will largely help us to tackle the 12 problem types that have come up in JEE Mains in last 3 years.

So let us start with the 24 variations. What you can understand is that in any question of P&C, what can happen is the object type can change. So, there might be distinct objects like a b c d is distinct or there can be identical objects like a a a a. And then there can be mixture of both like two A's and 2 B's so this is one variation. Another variation that happens is either we are going to do selection or we are going to do arrangement or both of these things. Another variation that can happen is we can do this using all the objects like from ABCD I can take two objects or I can take all the objects, either you can take some or you can take all of them. And ultimately repetition, so whether the repetition is allowed or not in that in that question. So, you can see that if I apply my principle of multiplication on it this becomes 3 into 2 into 2 into 2. That is equal to 24. So, what happens is the same problem I can take and based on these different ideas I can write 24 variations to the same problem.

So now let us take one very simple problem and see how these 24 variations work out. So, the problem is very simple. We have been given five distinct letters a b c d e. The first question is we have to take all five and make words such that there is no repetition happening. What will be the ways to do it?

In this case what you can use is either nPr so what you can do is you can write the answer directly by saying 5P5, which is nothing but 5 factorial. Or you can use nCr into R

Factorial, which is also very popular method. So, what I can do here is I can write 5C5 into 5 factorial. This can also be done or what I can use is nothing but box method. So, what I can do is I can make boxes and then fill them sequentially so this is nothing but principle of

counting and ultimately, we stop putting these arrows and we just make boxes so what I can do is now if I have to make a word in the arrangement. The order in which the letters are coming is important, so we can apply here principle of counting and we can write the first box can be filled in five ways since a distinct letter is getting used. Next can be filled in four then three then two then one way.

So, if I multiply it what I get is nothing but 5 factorial. So, from here also I'm getting the answer as 5 factorial. In this method in this type of question you can apply any of the three methods and you will get the right answer.

Now if I change this question a little bit and say that the question is almost same but instead of these five letters words, I need to make three letter words, the repetition is still not allowed. So, what I can do is again I can write my answer by using nPr or 5P3. Or I can write my answer as 5C3 into 3 factorial, which is also fine. Or I can use my box method and now I have to fill three different boxes, so answer becomes 5 into 4 into 3.

Now the next problem says that you have to again make five letter words but now the repetition of letters is allowed. So earlier I could not write a a a a a but now in this case I can make words in which the letters are also repeating. How many such words are possible. If I do this again now only one method is permissible that is nothing but our box method because whenever the repetition will happen nPr and nCr will not be valid. This is a very important and significant statement that you need to remember. All I can use is this basic principle of counting and here if I apply the first box can be filled is nothing but again five ways and since you can see that I can repeat the letter the next can also be filled in five ways and again five so our answer in this case becomes nothing but 5 to the power 5.

Now this similar approach can be used in having these three letters where repetition is allowed. So again, we can make three boxes and the answer comes out to be 5 into 5 into 5. Again, point to be noted we cannot use nPr and nCr. See usually the students have this habit of throwing nPr and nCr to The Faculty or the question maker whenever they see a problem instead trying and developing a very analytical way of applying methods. What method can be applied where and more importantly how are you thinking and using these methods to get to the answer.

Let us move to the second part that is combination or selection. Let us see that we have five types of chocolates you can assume it we have ABCDE five types of chocolates and my first question is if you have five types of chocolates, you need to choose five chocolates. You cannot repeat the chocolate. So, if you have five chocolates you can take only one of each type and you have to choose five chocolates, the first part becomes really simple. What you can do is you can apply nCr and write 5C5 which comes out to be one. From Common Sense also you can write the answer as one.

If I have to choose three chocolates then the answer will become 5C3. In the next case, which is very interesting, out of these five chocolates you can choose each one any number of times. As far as I'm choosing five chocolates, I can choose either also five or I can choose B also five or I can choose D also etc. Or I can take 3a and 2B and or anything similar. I can choose any kind of combination that I can think and so this becomes a very interesting question.

Now how to solve this question becomes very important. In such kind of questions, we will use something called as Beggar’s method. We will write a small number of chocolates that we are choosing in the small a as capital A. So, A + B + C + D + E equals five. You can understand this question like this - that your mother took you to the market and said that you can choose any kind of chocolate as far as you are choosing only five chocolates and there's five different boxes of chocolates placed in front of you. You can put hand in one of those boxes and take any number of chocolates as far as you are choosing.  Five everything is fine and she will not get worried of your health so this is what is happening in this question. What we are saying is A is what do I chose from box a. B is something that I have chosen from box b and so forth. Beggar's method says is that the answer can be directly written as five whatever number of quantities we are writing on this side to again they are five subtract one and C5. Whatever we wrote here we have also written here again five whatever is on the right-hand side how many quantities are there on the left-hand side minus one and C five whatever is on the right-hand side this becomes a standard method known as Beggar's method and the answer of this question becomes 9C5

We move along in the video with all such variants and then question types of each variant.

Now if we look into the weightage of this chapter, generally there is one question from the chapter and that can get very simple to solve if you apply your thought process as explained in the video.

Problem Types discussion

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