Sample Test Question And Answers For Permutations And Combinations Pdf

Introduction

Understanding permutations and combinations is essential for anyone preparing for competitive exams, interviews, or entrance tests that assess mathematical reasoning. This article provides a comprehensive set of sample test questions and answers that can be saved as a PDF for quick revision. By working through these examples, you will master the core concepts, learn efficient problem‑solving techniques, and gain confidence in tackling any permutation‑combination question that appears on your exam Worth keeping that in mind..

Why Permutations and Combinations Matter

• Real‑world relevance – From arranging books on a shelf to forming committees, the principles of ordering (permutations) and selection (combinations) appear in everyday decision‑making.
• Exam frequency – Questions involving nPr (permutations) and nCr (combinations) are a staple in quantitative sections of tests such as GRE, GMAT, CAT, and various government exams.
• Foundation for probability – Many probability problems rely on the ability to count arrangements and selections accurately.

Core Concepts and Formulas

1. Permutations (Ordered Arrangements)

The number of ways to arrange r objects out of n distinct objects is denoted as nPr and calculated by:

[ \text{nPr} = \frac{n!}{(n-r)!} ]

Key point: When r = n, the formula simplifies to n!, the total number of possible orders.

2. Combinations (Unordered Selections)

The number of ways to choose r objects from n distinct objects without regard to order is denoted as nCr and given by:

[ \text{nCr} = \frac{n!}{r!(n-r)!} ]

Key point: nCr = nPr / r!, because dividing by r! removes the effect of ordering.

3. Factorial Notation

The symbol ! denotes the factorial of a non‑negative integer n, defined as:

[ n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1 ]

Special case: 0! = 1 by definition, which makes many formulas work smoothly.

Sample Test Questions and Answers

Below are five representative questions that cover a range of difficulty levels. Each question is followed by a step‑by‑step solution, with the final answer highlighted in bold And that's really what it comes down to..

Question 1 – Basic Combination

Question: In how many ways can a committee of 3 members be selected from a group of 10 people?

Solution:

1. Identify that order does not matter → use nCr.
2. Apply the formula:

[ \text{10C3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!,7!} ]

3. Cancel common terms:

[ \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120 ]

Answer: 120 ways.

Question 2 – Permutation of Distinct Items

Question: How many different 4‑letter “words” can be formed using the letters of the word “MATH” without repetition?

Solution:

1. All letters are distinct, and we need to arrange 4 of them → use 4P4.
2. Since we are arranging all available letters,

[ \text{4P4} = 4! = 24 ]

Answer: 24 possible words Simple, but easy to overlook. No workaround needed..

Question 3 – Mixed Permutation and Combination

Question: A class has 8 boys and 6 girls. If a team of 5 students must contain exactly 3 boys, in how many ways can the team be formed?

Solution:

1. Choose 3 boys from 8 → 8C3.
2. Choose the remaining 2 students (girls) from 6 → 6C2.
3. Multiply the two selections (Fundamental Counting Principle):

[ \text{8C3} \times \text{6C2} = \frac{8!}{3!5!} \times \frac{6!}{2!4!} ]

4. Compute each part:

[ \text{8C3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 ]

[ \text{6C2} = \frac{6 \times 5}{2 \times 1} = 15 ]

5. Final multiplication:

[ 56 \times 15 = 840 ]

Answer: 840 ways It's one of those things that adds up. Less friction, more output..

Question 4 – Permutation with Repetition

Question: How many distinct arrangements can be made from the letters of the word “BOOKKEEPER”?

Solution:

1. Count total letters: 10.
2. Identify repetitions: B(1), O(2), K(2), E(3), P(1), R(1).
3. Use the permutation formula for multisets:

[ \frac{10!}{1!,2!,2!,3!,1!,1!} ]

4. Simplify:

[ \frac{3628800}{2 \times 2 \times 6} = \frac{3628800}{24} = 151200 ]

Answer: 151,200 distinct arrangements.

Question 5 – Advanced Combination with Constraints

Question: From a set of 12 numbers, how many subsets of size 4 contain at least one even number? (Assume the set contains 6 even and 6 odd numbers.)

Solution:

1. Use the complementary counting method: total subsets of size 4 minus subsets with no even numbers (i.e., all odd).
2. Total subsets:

[ \text{12C4} = \frac{12!}{4!8!} = 495 ]

3. Subsets with only odd numbers (choose 4 from the 6 odd):

[ \text{6C4} = \frac{6!}{4!2!} = 15 ]

4. Subtract:

[ 495 - 15 = 480 ]

Answer: 480 subsets

Question 6 – Circular Permutation

Question: Ten people are to be seated around a round table. How many different seating arrangements are possible if two particular people, Alice and Bob, must not sit next to each other?

Solution:

1. Total circular arrangements of 10 distinct people:

[ (10-1)! = 9! = 362{,}880 ]

(Fix one person as a reference point to eliminate rotational symmetry.)

2. Count the “bad’’ arrangements where Alice and Bob are adjacent.
   • Treat Alice & Bob as a single block. The block plus the remaining 8 individuals give 9 “objects’’ to arrange around the circle.
   • Circular arrangements of 9 objects:

[ (9-1)! = 8! = 40{,}320 ]

• Inside the block, Alice and Bob can be ordered in 2 ways (Alice‑Bob or Bob‑Alice).

[ \text{Bad arrangements} = 2 \times 8! = 2 \times 40{,}320 = 80{,}640 ]

3. Subtract the bad cases from the total:

[ \text{Good arrangements} = 9! - 2 \times 8! = 362{,}880 - 80{,}640 = 282{,}240 ]

Answer: 282,240 seating arrangements in which Alice and Bob are not neighbours Simple, but easy to overlook..

Question 7 – Permutations with Identical Objects

Question: A bakery sells cupcakes in boxes of 6. Each box must contain exactly 2 chocolate, 2 vanilla, and 2 strawberry cupcakes. If the cupcakes are indistinguishable within each flavour, how many distinct ways can the bakery arrange the cupcakes in a single box?

Solution:

The problem reduces to arranging a multiset with repetitions:

[ \frac{6!}{2!,2!,2!}= \frac{720}{8}=90 ]

Answer: 90 distinct arrangements Took long enough..

Question 8 – Combination with a “At Most” Condition

Question: A committee of 7 is to be formed from 9 engineers and 5 managers. The committee may contain at most 3 managers. In how many ways can the committee be formed?

Solution:

We sum the possibilities for 0, 1, 2, and 3 managers Easy to understand, harder to ignore. Less friction, more output..

Add them up:

[ 36 + 420 + 1{,}260 + 1{,}260 = 2{,}976 ]

Answer: 2,976 possible committees Worth keeping that in mind..

Question 9 – Permutations with a Fixed Position

Question: In how many ways can the letters of the word “PROBLEM” be arranged if the letter ‘P’ must remain in the first position?

Solution:

1. Fix ‘P’ at the first slot.
2. The remaining six distinct letters (R, O, B, L, E, M) can be arranged freely:

[ 6! = 720 ]

Answer: 720 arrangements No workaround needed..

Question 10 – Inclusion–Exclusion Principle

Question: A deck contains 52 cards. How many 5‑card hands contain at least one ace or at least one king?

Solution:

Let

• (A) = hands containing at least one ace.
• (K) = hands containing at least one king.

We need (|A \cup K| = |A| + |K| - |A \cap K|).

1. Count (|A|):
   Total hands: (\binom{52}{5}).
   Hands with no ace: (\binom{48}{5}).
   Hence (|A| = \binom{52}{5} - \binom{48}{5}).

2. Count (|K|):
   Analogously, (|K| = \binom{52}{5} - \binom{48}{5}) (since there are also 4 kings) It's one of those things that adds up..

3. Count (|A \cap K|):
   Hands containing at least one ace and at least one king.
   Compute the complement: hands with no ace or no king And it works..

   • No ace: (\binom{48}{5}).
   • No king: (\binom{48}{5}).
   • No ace and no king (i.e., exclude all 8 cards): (\binom{44}{5}).

   By inclusion–exclusion on the complement:

[ \text{Hands missing at least one of the two ranks} = \binom{48}{5} + \binom{48}{5} - \binom{44}{5} ]

Therefore

[ |A \cap K| = \binom{52}{5} - \bigl[ \binom{48}{5} + \binom{48}{5} - \binom{44}{5} \bigr] ]

4. Combine:

[ |A \cup K| = 2\bigl[\binom{52}{5} - \binom{48}{5}\bigr] - \Bigl{ \binom{52}{5} - \bigl[ 2\binom{48}{5} - \binom{44}{5} \bigr] \Bigr} ]

Simplify:

[ |A \cup K| = \binom{52}{5} - \binom{44}{5} ]

5. Numerical evaluation:

[ \binom{52}{5}=2{,}598{,}960,\qquad \binom{44}{5}=1{,}086{,}008 ]

[ |A \cup K| = 2{,}598{,}960 - 1{,}086{,}008 = 1{,}512{,}952 ]

Answer: 1,512,952 five‑card hands contain at least one ace or at least one king.

Wrapping It Up

The ten examples above illustrate a spectrum of counting techniques that are the backbone of combinatorial problem‑solving:

• Basic combinations ((nC r)) for “choose‑without‑order’’ scenarios.
• Permutations for ordered selections, with or without repetition.
• Mixed cases that combine the two ideas, often resolved with the Fundamental Counting Principle.
• Special constraints – circular arrangements, forbidden adjacencies, identical objects, and “at most/at least’’ conditions – which require a tweak to the standard formulas or a complementary‑counting approach.
• Inclusion–exclusion, a powerful tool when multiple overlapping restrictions are present.

Mastering these patterns enables you to translate almost any discrete‑selection story into a clean mathematical expression, compute it efficiently, and, most importantly, verify that you’ve accounted for every possibility. Practice by altering the numbers or adding new constraints; the underlying logic stays the same, and with each variation your intuition for counting grows stronger.

Bottom line: Whether you’re assembling a sports team, arranging letters, or shuffling a deck of cards, the combinatorial toolbox—combinations, permutations, and the principles that connect them—gives you a reliable roadmap to the answer. Keep this guide handy, experiment with the formulas, and you’ll find that even the most complex counting puzzles become manageable, step by step Practical, not theoretical..