Class 11 Maths permutations and combinations: home tuition guide
Permutations and combinations is the chapter where many Class 11 students first find that Maths cannot be learnt by copying solutions.
This guide shows the order in which to learn the chapter, where marks are usually lost, and a weekly home routine that fixes it.
Why this chapter troubles Class 11 students
Most chapters give you a method and ask you to apply it. This chapter gives you a situation and asks you to decide the method. That is a new skill.
The formulas are short. The difficulty is in reading the question. A student who knows both formulas can still lose marks by using the wrong one.
Parents often say that the child knows the formula but still gets the answer wrong. This is usually the reason. The weakness is in reading and sorting the problem, not in memory.
The chapter also needs comfort with fractions and cancelling, because every answer involves factorials. Weak arithmetic shows up quickly here.
The concept ladder: five steps from counting to problems
Teach the chapter in this order. Do not move up a step until the student can explain the current one in their own words.
- Step 1: Fundamental principle of counting. If one job can be done in m ways and a second job in n ways, doing both, one after the other, can be done in m × n ways. This is the multiplication rule. If you may do either job but not both, you add the ways. This is the addition rule. Everything later is built on these two rules.
- Step 2: Factorial. The factorial of n, written n!, is the product 1 × 2 × 3 × … × n. It counts the ways to arrange n different things in a row. By definition, 0! = 1. Students should practise simplifying a fraction like 7!/5! without expanding everything.
- Step 3: Permutations. A permutation is an arrangement where order matters. The number of ways to arrange r things taken from n different things is nPr = n!/(n − r)!.
- Step 4: Combinations. A combination is a selection where order does not matter. The number of ways to choose r things from n is nCr = n!/(r!(n − r)!). Every selection of r things can be arranged in r! ways. That is why the combination formula has an extra r! in the denominator.
- Step 5: Mixed problems. Words with repeated letters, committees with conditions, numbers formed from digits, and at-least-one questions. These use steps 1 to 4 together.
Most students who struggle are stuck at step 1 or 2 but are already working at step 5. A good tutor goes back to the step where the understanding broke.
Permutation or combination: the question to ask first
Before writing any formula, ask yourself: if I swap two of the chosen things, is the result different? If yes, order matters, so it is a permutation. If no, it is a combination.
| Situation | Does order matter? | Use |
|---|---|---|
| Choosing a president and a secretary from a group | Yes, the two posts are different | Permutation |
| Choosing two members for a committee | No, both have the same role | Combination |
| Forming a 3-digit number from given digits | Yes, a number and its reverse are different numbers | Permutation |
| Choosing 3 books to carry from a shelf | No, the same three books are the same choice | Combination |
| Arranging the letters of a word | Yes | Permutation |
Ask the student to fill a table like this on their own for ten new situations. They should only classify, not solve. This one exercise removes many wrong answers.
Three worked examples
Example 1 (permutation). How many 3-digit numbers can be made from the digits 1, 2, 3, 4, 5 if no digit is repeated?
The hundreds place can be filled in 5 ways. The tens place then has 4 choices left. The units place has 3. By the multiplication rule: 5 × 4 × 3 = 60. This is the same as 5P3 = 5!/2! = 60.
Example 2 (repeated letters). How many different arrangements can be made from the letters of a six-letter word in which one letter appears 3 times and another letter appears 2 times?
All six letters could be arranged in 6! ways. Repeated letters cannot be told apart, so we divide by 3! and by 2!: 6!/(3! × 2!) = 720/12 = 60.
Example 3 (combination with a condition). A committee of 3 is to be chosen from 5 men and 4 women. It must have at least one woman. How many committees are possible?
Count all committees first: 9C3 = 84. Now count the committees with no woman, that is, all men: 5C3 = 10. Subtract: 84 − 10 = 74.
Subtracting the unwanted cases from the total is much shorter than adding the cases of one woman, two women and three women. Students should look for it whenever they see at least one.
Common errors a tutor looks for
- Using nPr when order does not matter. Fix: ask the swap question, every time, in writing.
- Forgetting that 0! = 1. This breaks cases such as nCn and nC0, both of which equal 1.
- Not dividing for repeated letters. Always check whether any letter repeats before using n!.
- Ignoring the condition on the first place. In number problems, the first digit cannot be 0 if a proper number is needed. Fill the restricted place first.
- Expanding factorials fully. Writing out 10! wastes time and invites mistakes. Cancel the common part instead.
- Missing the identity nCr = nCn−r. It turns 50C48 into the easy 50C2.
How one-to-one home tuition helps in this chapter
Counting problems cannot be learnt by watching. The student must try, go wrong, and explain why. In a one-to-one class at home, the tutor can see the exact line where the thinking turned wrong. In a large group, that moment passes unnoticed.
Great Home Tutors offers home tuition only. There is no coaching centre and no group batch. You give the class, subject, board and your locality, and a person from the team calls you back to understand the student. We tell you plainly whether a tutor is verified by us or not yet, and there is a first class to see the fit before anything is fixed. This works for families in Dwarka and in Gurugram.
In this chapter the tutor should do three things: go back to the step where understanding broke, make the student say the reason aloud before writing a formula, and give mixed problems so the student cannot guess the method from the page.
A weekly routine that works at home
Adjust this to the pace of the school.
- Day 1, with the tutor: learn one step of the ladder with two or three solved examples.
- Day 2, alone: redo the same examples without looking, then solve five similar NCERT problems.
- Day 3, alone: classify ten situations as permutation or combination, without solving.
- Day 4, with the tutor: the tutor checks the wrong answers and asks the student to explain each mistake.
- Day 5, alone: a short mixed set of problems, timed, to build exam speed.
- Weekend: keep one notebook page for the week's mistakes. Read it before the next class.
Parents do not need to know the chapter to help. Ask the student to teach you one problem. If they can explain why they chose the formula, they have understood it.
Key takeaways
- Learn in order: counting rule, factorial, permutations, combinations, mixed problems.
- Decide the type first. Ask whether swapping two chosen things changes the result.
- For at-least-one questions, try the total minus the unwanted cases first.
- Check for repeated letters, the rule 0! = 1 and restricted first places before finalising an answer.
- Practise a little every day and keep a record of mistakes. Progress depends on regular work.
Frequently asked questions
Is permutations and combinations difficult in Class 11?
The formulas are short, but the chapter asks the student to choose the method. Students who practise sorting problems by type usually find it manageable.
Should my child memorise the formulas or understand them?
Both, but understanding comes first. If the student knows that a permutation of r things equals a combination multiplied by r!, the two formulas are easy to remember and hard to mix up.
How many problems should my child solve each day?
Quality matters more than number. A few problems solved carefully, with the reason written for each, are worth more than many copied quickly. Start with the NCERT examples and exercises.
When should we start home tuition for this chapter?
Start when the school begins the chapter, or as soon as the student begins to guess formulas. Early help is easier than repairing weak basics later, because counting ideas are used again in later chapters.
Can the tutor teach at our home in Dwarka or Gurugram?
Great Home Tutors arranges one-to-one tuition at your home. Use the request form, give the class, subject, board and your locality, and a person from the team will call you back.
What to do next. Try it yourself on the page below.
Great Home Tutors — home tuition at your home · 4AB →
