Permutations and Combinations is about counting without listing. When the NDA paper asks “in how many ways”, you rarely have time to write out every option. Instead you use two clean tools: permutations for ordered arrangements and combinations for unordered selections. Master the difference and you bag 2–4 quick marks every year.
Why This Topic Is a Scoring Goldmine
Almost every NDA Mathematics paper carries 2 to 4 questions straight from Permutations and Combinations, plus a few more hidden inside Probability. The good news is that the questions are formula-driven and short — once you spot whether order matters, the rest is plug-and-play.
This chapter also builds the foundation for Binomial Theorem and Probability, so the time you invest here pays off three times over. In Probability, almost every “favourable outcomes divided by total outcomes” calculation is really a counting problem in disguise — and counting is exactly what this chapter teaches.
What makes the topic friendly is that the questions are short and self-contained. You do not need long calculations or memorised tables; you need a clear head and the right formula. Many toppers treat these as “guaranteed marks” that they finish in the first ten minutes of the paper, leaving more time for the harder calculus and trigonometry questions later.
The single most important question to ask in every problem is: Does the order of the items matter? If yes → permutation. If no → combination. Everything else follows from that one decision.
The Fundamental Principle of Counting
This is the engine that powers the whole chapter. There are two parts.
Multiplication principle (AND)
If one job can be done in m ways and, after it, a second independent job in n ways, then both together can be done in m × n ways. Use it when events happen one after another (this AND then that).
Addition principle (OR)
If a job can be done by method A in m ways OR by method B in n ways, and the two methods cannot happen together, the total is m + n ways. The word “or” in everyday language is the clue that you should be adding rather than multiplying.
Every permutation and combination formula you will meet is just a neat shortcut for repeated use of these two principles. If you ever forget a formula in the exam, you can rebuild the answer from scratch by reasoning place-by-place with the multiplication principle. That is a safety net worth keeping.
A student can travel from Delhi to Agra by 3 trains or 2 buses, and from Agra to Jaipur by 4 trains. In how many ways can the full journey be made?
Factorial Notation You Must Know
The factorial of a positive integer n, written n!, is the product of all positive integers up to n.
n! = n × (n − 1) × (n − 2) × … × 2 × 1
0! = 1 (by definition)
n! = n × (n − 1)!
Quick values worth memorising: 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040.
Factorials are defined only for whole numbers. There is no such thing as (−3)! or (2.5)! at this level. Also, never write 0! = 0 — it equals 1, and forgetting this breaks many formulas.
Permutations: When Order Matters
A permutation is an arrangement of objects in a definite order. Choosing a President and a Vice-President from a group is a permutation, because who gets which post matters.
Number of permutations of n distinct objects taken r at a time:
nPr = n! ÷ (n − r)!
Special cases: nPn = n! and nP0 = 1.
For example, the number of ways to arrange 3 of 5 books on a shelf is 5P3 = 5! ÷ 2! = 120 ÷ 2 = 60.
Arranging all n objects
If you arrange every object, the count is simply n!. Five friends in a row = 5! = 120 ways. Notice this is just 5P5, because n! ÷ 0! = n!.
A good way to understand nPr is to imagine filling r empty boxes from left to right. The first box has n choices, the next has (n − 1) choices because one object is used up, then (n − 2), and so on for r boxes. Multiplying these together gives exactly n! ÷ (n − r)!. Seeing the formula this way means you never have to blindly memorise it.
Repetition, Identical Objects and Forming Numbers
When objects can repeat
If repetition is allowed, each of the r places can be filled by any of the n objects, giving nr arrangements. Example: the number of 3-digit codes using digits 0–9 with repetition is 103 = 1000.
When some objects are identical
Permutations of n objects where p are alike of one kind, q alike of another:
n! ÷ (p! × q! × …)
The word BALLOON (7 letters, with L twice and O twice) can be arranged in 7! ÷ (2! × 2!) = 5040 ÷ 4 = 1260 ways.
Forming numbers and words
To form 4-digit numbers from 1, 2, 3, 4, 5 with no repetition, fill four boxes: 5 × 4 × 3 × 2 = 120, which equals 5P4. If 0 is allowed, the first box cannot be 0, so count that place separately, then fill the rest.
For “always together” problems, tie the items into one block and arrange. For “never together”, count total arrangements and subtract the “together” ones. And always lock the most restricted position (like the units digit for divisibility) before the free ones.
Circular Permutations
When objects are placed around a circle (people around a round table, beads in a ring), there is no fixed starting point, so we fix one object and arrange the rest.
Circular arrangements of n distinct objects = (n − 1)!
If clockwise and anticlockwise look the same (like a garland or necklace) = (n − 1)! ÷ 2
So 6 people around a round table can be seated in (6 − 1)! = 5! = 120 ways. But 6 differently coloured beads in a necklace form only 5! ÷ 2 = 60 distinct rings.
Do not use n! for a circular table. Rotating everyone by one seat is the same arrangement, which is exactly why we divide out one position and get (n − 1)!.
Combinations: When Order Does Not Matter
A combination is a selection of objects where order is irrelevant. Picking a team of 3 players from 5 is a combination — the team is the same whatever sequence you pick them in.
Number of combinations of n distinct objects taken r at a time:
nCr = n! ÷ [r! × (n − r)!]
Relation to permutations: nPr = nCr × r!
Choosing 3 of 5 players: 5C3 = 5! ÷ (3! × 2!) = 120 ÷ 12 = 10 teams.
The link nPr = nCr × r! is worth understanding deeply. First you select r objects (that is nCr), and then you arrange those r objects among themselves (that is r!). So a permutation is simply a combination that has been ordered afterwards. If you ever mix the two up, this relation lets you correct the count instantly by multiplying or dividing by r!.
Handy properties of nCr
These shortcuts save real time in the exam.
nCr = nC(n − r) (symmetry)
nC0 = nCn = 1
nCr + nC(r − 1) = (n+1)Cr (Pascal’s rule)
If nCx = nCy, then either x = y or x + y = n.
The symmetry property is a real time-saver: instead of computing 50C48, use 50C2 = (50 × 49) ÷ 2 = 1225.
The total number of ways to select at least one object from n distinct objects is 2n − 1 (each object is either chosen or not, minus the empty case).
Permutation or Combination? A Quick Decision Guide
Most mistakes in this chapter come from choosing the wrong tool. Use these signal words.
Order matters → Permutation
- Arrange, line up, seat in a row
- Form numbers, words or codes
- Assign distinct posts (captain, secretary)
- Rank or schedule positions
Order does not matter → Combination
- Select, choose, pick a group or committee
- Form a team where roles are equal
- Draw cards or balls as a set
- Choose questions to attempt
“Arrangement” almost always means permutation; “selection” or “group” almost always means combination. Read the verb in the question.
A Mixed Worked Example
Many NDA questions combine selection and arrangement. Here is the standard two-step approach.
From 7 men and 4 women, a committee of 5 is to be formed with exactly 3 men and 2 women. In how many ways, and how many of these have the 5 members then seated in a row?
The key habit: select first, then arrange. Mixing the two steps is the commonest source of wrong answers.
Mistakes That Cost Easy Marks
- Confusing P and C: using nPr for a “selection” problem inflates the answer by a factor of r!.
- Ignoring repetition rules: “digits may repeat” versus “all digits different” give completely different counts.
- Forgetting the zero case: in number-formation, a number cannot start with 0, so handle the leading place separately.
- Adding when you should multiply: sequential independent choices use AND (multiply), not OR (add).
For forming 3-digit numbers from 1–5 with no repetition the answer is 5P3 = 60. But if the digits were 0–5, the first place can take only 5 values (no leading 0), so it becomes 5 × 5 × 4 = 100, not 6P3.
Previous-Year Style Practice
Try this in under a minute before reading the solution.
Q. In how many ways can the letters of the word SUNDAY be arranged so that the arrangement begins with S and does not end with Y?
Answer: Fix S in the first place. The remaining 5 letters (U, N, D, A, Y) fill 5 places: total = 5! = 120. From these, subtract arrangements ending with Y: fix S first and Y last, the middle 4 letters give 4! = 24. So valid arrangements = 120 − 24 = 96.
For “does not” or “at least” conditions, the fastest route is usually total minus the unwanted cases rather than counting the wanted cases directly.
Quick Revision
- Order matters → permutation (nPr = n!/(n−r)!); order does not → combination (nCr = n!/[r!(n−r)!]).
- Multiplication principle for AND (sequential), addition principle for OR (alternative).
- 0! = 1; n distinct objects in a row = n!; in a circle = (n−1)!.
- Identical objects: divide by the factorials of each repeated group.
- nCr = nC(n−r); at least one selection from n objects = 2n − 1.
- For mixed problems: select first, then arrange.
Frequently asked questions
What is the basic difference between a permutation and a combination?
A permutation counts ordered arrangements, so AB and BA are different. A combination counts unordered selections, so AB and BA are the same single group. If the question says arrange, the order matters; if it says select or choose, order usually does not.
Why is 0! equal to 1 and not 0?
0! is defined as 1 so that key formulas like nCr = n!/[r!(n-r)!] and nPn = n! stay consistent. For example nCn must equal 1, and that only works if 0! = 1. Treat it as a fixed definition to memorise.
How do I quickly decide whether to multiply or add counts?
Use multiplication when choices happen one after another and all are needed (this AND then that). Use addition when there are separate, mutually exclusive ways to do the same job (this way OR that way).
How are circular permutations different from row arrangements?
In a row, n distinct objects give n! arrangements. In a circle there is no fixed starting point, so one object is fixed and the rest arranged, giving (n-1)!. For garlands or necklaces, where clockwise and anticlockwise are identical, divide further by 2.
How many questions from this topic appear in the NDA exam?
Permutations and Combinations typically contributes about 2 to 4 direct questions in NDA Mathematics, and it also underpins several Probability questions. Because the problems are short and formula-based, it is one of the most reliable scoring areas.
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