Quantitative Aptitude · Permutations and Combinations

Combinations (nCr) and Its Properties

Updated 1 October 2026 · Fact-checked

A combination is a selection of r objects from n distinct objects where order does not matter. Its count is nCr = n! ÷ (r! × (n − r)!). To solve questions, use nCr = nC(n − r), Pascal's relation, and for nCx = nCy, set x = y or x + y = n. Check that answers are valid.

Understand Combinations (nCr) and Its Properties

A combination is a way of choosing objects when the order of choosing does not matter. Picking a team of 3 from 10 students is a combination: {Asha, Ravi, Meena} is the same team as {Meena, Asha, Ravi}.

Why the formula looks the way it does: if order did matter, you would count nPr = n! ÷ (n − r)! arrangements. But every group of r objects gets counted r! times, once for each way of ordering it. So you divide by r! and get nCr = n! ÷ (r! × (n − r)!).

The key property is nCr = nC(n − r). Choosing r objects to take is the same as choosing n − r objects to leave behind. That is why 20C18 is just 20C2. Use it to keep numbers small.

Pascal's relation says nCr + nC(r − 1) = (n + 1)Cr. Think of one special object among n + 1. Either you pick it (then choose r − 1 from the other n) or you do not (then choose r from the other n). Add the two cases.

For equations, if nCx = nCy, then either the two numbers are equal (x = y) or they are complements (x + y = n). Always check that the final n and r are whole numbers with 0 ≤ r ≤ n.

Key formulas to remember

Combination formula: nCr = n! ÷ (r! × (n − r)!); Valid for whole numbers n and r with 0 ≤ r ≤ n.
Link with permutations: nCr = nPr ÷ r!; Divide by r! to remove the order.
Complementary rule: nCr = nC(n − r); Use it when r is more than n ÷ 2 to shrink the calculation.
Special values: nC0 = nCn = 1; nC1 = n; Only one way to choose none or all; n ways to choose one.
Pascal's relation: nCr + nC(r − 1) = (n + 1)Cr; Adds two neighbouring values of the same n to give the next n.
Ratio of consecutive terms: nCr ÷ nC(r − 1) = (n − r + 1) ÷ r; Very useful for equations and for finding n or r from ratios.
Reduction formula: nCr = (n ÷ r) × (n − 1)C(r − 1); Needs r ≥ 1.
Equal combinations: If nCx = nCy, then x = y or x + y = n; Check both cases. Reject values that are not whole numbers or exceed n.
Sum of all combinations: nC0 + nC1 + ... + nCn = 2ⁿ; Selecting at least one object from n distinct objects gives 2ⁿ − 1 ways.

How to solve Combinations (nCr) and Its Properties questions

Use this order for any nCr question, whether it asks for a value, an equation or a proof-type identity.

1Write down n and r clearly and confirm the objects are distinct and order does not matter.
2If r is greater than n ÷ 2, replace r by n − r using nCr = nC(n − r).
3Expand only as many factors as needed: nCr = [n × (n − 1) × ... (r factors)] ÷ r!. Cancel before multiplying.
4For an equation nCx = nCy, write both cases: x = y and x + y = n. Solve each.
5For sums of two nCr terms with the same n and neighbouring r, apply Pascal's relation to combine them.
6For ratio or unknown n, use nCr ÷ nC(r − 1) = (n − r + 1) ÷ r or the reduction formula, then solve the simple equation.
7Check validity: n and r must be whole numbers with 0 ≤ r ≤ n. Reject anything else.
8Match your value with the options and mark it.

Quickest way: Complement, cancel and case-split

When to use it: Use in the MCQ round when you see a numeric nCr, an equation with nCx = nCy, or a sum of two adjacent nCr terms.

Numeric value: shrink r with nCr = nC(n − r), then multiply r numbers from the top and divide by r! while cancelling. For example 12C10 = 12C2 = (12 × 11) ÷ 2 = 66.
Equation nCx = nCy: write x + y = n at once. Solve it, then separately try x = y. Test every answer in the options for validity.
Special case nCr = nC(r + 1): r = r + 1 is impossible, so only the complement case applies. The two indices add to n, so n = 2r + 1. This means n must be odd.
Adjacent terms: nCr + nC(r + 1) = (n + 1)C(r + 1). Do not compute either term.
If the algebra gets long (more than about 90 seconds), put in an option value for n or r and test. Skip if still unclear, since a wrong answer costs 0.25.

Common mistakes in Combinations (nCr) and Its Properties

Using only x + y = n and missing the case x = y (or the reverse).
Students remember the complement rule and forget that equal indices also give equal values.
Fix: Always write both cases first, solve each, and then test validity.
Accepting a non-integer or too large value, such as r = 1.5.
The algebra gives an answer and students stop there.
Fix: Check that n and r are whole numbers and that 0 ≤ r ≤ n. Reject anything else.
Computing nPr and forgetting to divide by r!.
The permutation and combination formulas look alike.
Fix: Ask: does order matter? If not, divide by r!. Sanity check: nCr is never more than nPr.
Writing nCr + nC(r − 1) = nC(r + 1) or = nC(2r − 1).
Students half-remember Pascal's relation.
Fix: The answer has n + 1 on top and the larger r: nCr + nC(r − 1) = (n + 1)Cr.
Expanding factorials in full for big numbers like 30C28.
Students miss the complement rule and waste time.
Fix: Switch to 30C2 = (30 × 29) ÷ 2 = 435 before doing any multiplication.
Treating 0! as 0 and getting nCn or nC0 wrong.
Zero feels like it should give zero.
Fix: Remember 0! = 1, so nC0 = nCn = 1.

Worked examples

Example 1

If nC8 = nC12, find the value of nC2. Options: (a) 380 (b) 190 (c) 20 (d) 210

Show the solution

Here x = 8 and y = 12. Case x = y is impossible since 8 ≠ 12.
So the complement case holds: n = 8 + 12 = 20.
Now nC2 = 20C2 = (20 × 19) ÷ 2.
= 380 ÷ 2 = 190.

Answer: (b) 190

Example 2

If 15C(3r) = 15C(r + 3), what is the whole-number value of r? Options: (a) 3 (b) 4 (c) 5 (d) 6

Show the solution

Case 1: 3r = r + 3 gives 2r = 3, so r = 1.5. This is not a whole number, so reject it.
Case 2: 3r + (r + 3) = 15 gives 4r = 12, so r = 3.
Check: 15C9 and 15C6. Since 9 + 6 = 15, they are equal. Both indices lie between 0 and 15, so r = 3 is valid.

Answer: (a) 3

Example 3

What is the value of 10C4 + 10C5? Options: (a) 504 (b) 330 (c) 462 (d) 252

Show the solution

Both terms have n = 10 and neighbouring r values (4 and 5). Use Pascal's relation: nCr + nC(r − 1) = (n + 1)Cr, with r = 5.
So 10C5 + 10C4 = 11C5.
11C5 = (11 × 10 × 9 × 8 × 7) ÷ (5 × 4 × 3 × 2 × 1) = 55440 ÷ 120 = 462.
Check: 10C4 = 210 and 10C5 = 252, and 210 + 252 = 462.

Answer: (c) 462

Exam tips

Questions on this topic are usually one-step or two-step. If you are still expanding factorials after a minute, you have missed a shortcut such as the complement rule or Pascal's relation.
In equation questions, the options often include a value from the wrong case or an invalid value. Test the case x = y and the case x + y = n before choosing.
Always check n ≥ r. A tempting option may give r larger than n, which makes nCr meaningless.
Memorise nC0 = nCn = 1, nC1 = n, and nC2 = n(n − 1) ÷ 2. They appear again and again.
With negative marking of 0.25 per wrong answer, skip a question if you cannot see the method quickly. Come back after finishing the sure ones.

Practice questions from Permutations and Combinations

Combinations (nCr) and Its Properties: frequently asked questions

What is the formula for nCr in CA Foundation?

nCr = n! ÷ (r! × (n − r)!), where n and r are whole numbers and 0 ≤ r ≤ n. It gives the number of ways to choose r objects from n distinct objects when order does not matter.

How do I solve nCx = nCy?

Either x = y, or x + y = n. Solve both cases, then reject any value that is not a whole number or that makes r greater than n. This works because choosing x objects is the same as leaving out n − x.

How do I prove nCr = nC(n − r)?

Write nC(n − r) = n! ÷ ((n − r)! × (n − (n − r))!) = n! ÷ ((n − r)! × r!). This is exactly the formula for nCr. In words, choosing r objects to take is the same as choosing n − r to leave.

What is Pascal's relation?

nCr + nC(r − 1) = (n + 1)Cr. It lets you add two neighbouring combinations of the same n in one step. It is often used to simplify sums in MCQs.

When should I use nPr and when nCr?

Use nPr when order matters, such as arranging people in a row. Use nCr when only the group matters, such as choosing a committee. Remember nCr = nPr ÷ r!.