Permutations and Combinations — Class 11 MCQs with Answers

Permutations and Combinations — important questions & MCQs with answers (Class 11 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Value of 5! is:

   A60B120C25D100
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   Answer: 120

   Explanation: 5! means 5×4×3×2×1 = 120, the product of all positive integers up to 5 — not 5×5 = 25, and not a partial product like 4×3×2×1 = 24.

2. Q2 of 6Easy

   C(n,r) =

   Ar!/n!Bn!/r!Cn!Dn!/(r!(n−r)!)
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   Answer: n!/(r!(n−r)!)

   Explanation: ⁿCᵣ = n!/(r!(n−r)!) — selections where order doesn't matter. Skipping the r! divisor gives ⁿPᵣ instead, which overcounts each selection r! times.

3. Q3 of 6Medium

   P(5,2) =

   A60B10C20D5
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   Answer: 20

   Explanation: ⁿPᵣ = n!/(n−r)! counts ordered arrangements: P(5,2) = 5×4 = 20. Dividing by 2! as for combinations gives 10, wrong here since order matters.

4. Q4 of 6Medium

   Assertion (A): If ⁿC₉ = ⁿC₈, then n = 17. Reason (R): If ⁿCᵣ = ⁿCₛ with r ≠ s, then r + s = n.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: R is true, because ⁿCᵣ = ⁿCₙ₋ᵣ, so equal values with r ≠ s need s = n – r. Using R with r = 9 and s = 8 gives n = 9 + 8 = 17, so A is true and R explains it.

5. Q5 of 6Hard

   Number of arrangements of n distinct objects in a circle:

   An!B(n−1)!Cn!/2Dn²
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   Answer: (n−1)!

   Explanation: Circular arrangements of n distinct objects give (n−1)!, not n!, because fixing one object's seat removes the rotational duplicates a linear count would include.

6. Q6 of 6Hard

   Assertion (A): The number of ways of choosing 4 cards from a pack of 52 playing cards so that all four are face cards is 495. Reason (R): Selecting r objects out of n objects is the same as rejecting n – r objects, so ⁿCᵣ = ⁿCₙ₋ᵣ.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true but R is not the correct explanation of A

   Explanation: A is true: there are 12 face cards, and 4 of them can be chosen in ¹²C₄ = 495 ways. R is true too, but rejecting objects is not how A is counted, so R does not explain A.

Permutations and Combinations notes: key concepts (Class 11 Mathematics)

Counting problems ask how many ways an outcome can happen without listing them all. This chapter builds the fundamental principle of counting, then permutations for ordered arrangements and combinations for unordered selections, developing nPr and nCr, the identities linking them, and methods for repetition and restrictions.

Fundamental principle of counting

If one stage happens in m ways and the next independently in n ways, together they happen in m × n ways; multiply the choices at every stage.

Addition principle

When outcomes are mutually exclusive alternatives, add their counts rather than multiply; the guide is OR → add, AND → multiply the possibilities.

Factorial notation

n! = n × (n−1) × … × 2 × 1 with 0! defined as 1; it counts the arrangements of n distinct objects placed in a row.

Permutations (nPr)

An ordered arrangement of r objects chosen from n distinct objects; nPr = n!/(n−r)!, since order matters and each position is filled distinctly.

Combinations (nCr)

A selection of r objects from n where order is irrelevant; nCr = n!/[r!(n−r)!], found by dividing nPr by the r! orderings of each group.

Permutation–combination link

Each combination of r objects can be ordered in r! ways, so nPr = nCr × r!; this bridges selecting a group and then arranging it.

Permutations with repetition

When objects may repeat, r positions each filled n ways give nʳ arrangements; the factorial formula instead assumes distinct, non-repeating objects.

Arrangements of alike objects

n objects with p alike of one kind and q of another arrange in n!/(p! q! …) ways, dividing out the indistinguishable rearrangements.

Symmetry identity

nCr = nC(n−r): choosing the r to keep is the same as choosing the n−r to leave out, which cuts the work when r is close to n.

Pascal's identity

nCr + nC(r−1) = (n+1)Cr, the rule that generates Pascal's triangle and links coefficients in successive rows to one another.

Restricted arrangements

Handle constraints by fixing forced positions first, or treat objects that must stay together as one block, then arrange within that block.

Boundary values

nC0 = nCn = 1: there is exactly one way to choose nothing and one way to choose everything, which anchors the coefficient sequence.