Permutations & Combinations Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice permutations and combinations (combinatorics) with the most important counting tools: factorials and \(0!\), the fundamental counting principle (rule of product), permutations \(P(n,r)=\dfrac{n!}{(n-r)!}\) when order matters, combinations and binomial coefficients \(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\) when order does not matter, circular permutations (round-table seating), and classic counting applications like arrangements with repeated letters, bit strings, and polygon diagonals. If you want a refresher, click Start lesson to open a step-by-step guide with worked examples and quick checks.
Answer the question set and review your mistakes at the end.
How this permutations & combinations practice works
1. Take the practice set: answer the permutations, combinations, factorial, and counting questions below.
2. Open the lesson (optional): review the difference between order matters vs order does not matter, then learn the core formulas and patterns.
3. Retry: return to the question set and apply the right counting method immediately.
What you will learn in the permutations & combinations lesson
Counting foundations
Factorials \(n!\) and why \(0!=1\)
Fundamental counting principle (multiply choices step-by-step)
Rule of sum (add counts for disjoint cases)
Permutations (order matters)
Permutation formula \(P(n,r)=\dfrac{n!}{(n-r)!}\)
Fast reasoning: \(n\) choices, then \(n-1\), then \(n-2\), ...
Common traps: counting ordered arrangements when you meant to count selections
Combinations (order does not matter)
Binomial coefficient \(\binom{n}{r}\) and "n choose r" language
Relationship: \(P(n,r)=\binom{n}{r}\,r!\)
Symmetry: \(\binom{n}{r}=\binom{n}{n-r}\)
Classic applications
Circular permutations for round-table seating: \((n-1)!\)
Repeated elements (e.g., word arrangements): \(\dfrac{n!}{n_1!\,n_2!\cdots}\)
Bit strings, even/odd counting, and polygon diagonals via combinations
Purpose: Build a clear understanding of permutations and combinations so you can count arrangements and selections correctly using factorials, the fundamental counting principle, permutations \(P(n,r)\) (order matters), combinations \(\binom{n}{r}\) (order does not matter), plus common applications like circular permutations, repeated-letter arrangements, bit strings, and polygon diagonals.
Success criteria
Compute factorials and use \(0!=1\) correctly.
Apply the fundamental counting principle (multiply choices step-by-step).
Decide quickly: Does order matter? If yes, use permutations; if no, use combinations.
Use symmetry: \(\binom{n}{r}=\binom{n}{n-r}\), and connect permutations to combinations: \(P(n,r)=\binom{n}{r}\,r!\).
Count circular arrangements and arrangements with repeated elements.
Solve classic counting tasks: bit strings, even/odd restrictions, and diagonals.
Key vocabulary
Factorial: \(n!=n(n-1)(n-2)\cdots 2\cdot 1\) for \(n\ge 1\), and \(0!=1\).
Fundamental counting principle: if one step has \(a\) choices and another has \(b\) choices, total is \(ab\).
Permutation: an ordered arrangement. For \(r\) positions filled from \(n\) distinct items (no repetition): \(P(n,r)\).
Combination: an unordered selection. Choose \(r\) items from \(n\): \(\binom{n}{r}\).
Binomial coefficient: another name for \(\binom{n}{r}\), read “n choose r”.
Circular permutation: arrangements around a circle where rotations are considered the same: \((n-1)!\).
Repeated elements: if some items repeat, divide by factorials of repeat counts: \(\dfrac{n!}{n_1!\,n_2!\cdots}\).
Quick pre-check
Pre-check 1: What is \(5!\)?
Hint: \(5!=5\cdot 4\cdot 3\cdot 2\cdot 1\).
Pre-check 2: What is \(\binom{7}{0}\)?
Hint: \(\binom{n}{0}=1\) because there is exactly one way to choose nothing.
Factorials & Counting
Factorials and the fundamental counting principle
Learning goal: Count multi-step processes by multiplying choices, and recognize when factorials appear.
Key idea
The fundamental counting principle (rule of product) says: if a process has \(a\) choices for step 1, \(b\) choices for step 2, and \(c\) choices for step 3, then the total number of outcomes is \(a\cdot b\cdot c\).
A factorial counts arrangements of distinct items: \[ n!=n(n-1)(n-2)\cdots 2\cdot 1. \] There are \(n!\) ways to arrange \(n\) distinct objects in a line.
Worked example
Example: How many ways can you arrange the letters in “ABCD”?
There are 4 letters, all distinct: \[ 4!=4\cdot 3\cdot 2\cdot 1=24. \]
Try it
Try it 1: How many ways can 6 distinct books be arranged on a shelf?
Hint: Arranging 6 distinct items in a line gives \(6!\).
Try it 2: How many bit strings of length 4 are there in total?
Hint: Each bit has 2 choices. Use \(2\cdot 2\cdot 2\cdot 2 = 2^4\).
Summary
Use the rule of product to multiply choices across steps.
Use factorials to count arrangements of distinct items: \(n!\).
Permutations
Permutations: when order matters
Learning goal: Recognize “order matters” questions and compute \(P(n,r)\) correctly.
Key idea
A permutation is an ordered arrangement. If you fill \(r\) positions using \(n\) distinct items with no repetition, the count is: \[ P(n,r)=n(n-1)(n-2)\cdots(n-r+1)=\frac{n!}{(n-r)!}. \] A fast mental check: “first position has \(n\) choices, second has \(n-1\), …”.
Worked example
Example: What is \(P(5,2)\)?
Choose 2 items in order from 5 distinct items: \[ P(5,2)=5\cdot 4=20. \] (First pick: 5 choices, second pick: 4 choices.)
Try it
Try it 1: What is \(P(4,2)\)?
Hint: \(P(4,2)=4\cdot 3\).
Try it 2: What is \(P(10,1)\)?
Hint: Choosing 1 item in order from 10 distinct items gives 10 outcomes.
Summary
Use permutations when order matters.
\(P(n,r)=\dfrac{n!}{(n-r)!}\) counts ordered selections with no repetition.
Combinations
Combinations: when order does not matter
Learning goal: Recognize “order does not matter” questions and compute \(\binom{n}{r}\) (n choose r).
Key idea
A combination is an unordered selection. If you choose \(r\) items from \(n\) distinct items, the count is: \[ \binom{n}{r}=\frac{n!}{r!(n-r)!}. \] This is also called a binomial coefficient. A powerful connection: \[ P(n,r)=\binom{n}{r}\,r! \] (because each chosen group of \(r\) items has \(r!\) possible orders).
Worked example
Example: How many ways are there to choose \(3\) items from \(5\) distinct items?
Learning goal: Count round-table arrangements and arrangements when some items repeat.
Key idea
Circular permutations: When \(n\) distinct people sit around a round table and rotations are considered the same, the number of seatings is: \[ (n-1)!. \] We “fix” one person to remove rotational duplicates.
Repeated elements: If you arrange \(n\) items where some repeat (for example, a word with repeated letters), then divide by factorials of repeated counts: \[ \frac{n!}{n_1!\,n_2!\cdots}. \]
Worked example
Example: How many ways can you seat 5 people at a round table (rotations considered the same)?
\[ (5-1)!=4!=24. \]
Try it
Try it 1: How many ways can you seat 3 people around a round table (rotations considered the same)?
Hint: \((n-1)!\) with \(n=3\) gives \(2!=2\).
Try it 2: How many distinct arrangements of the letters in “MISS” are there?
Hint: “MISS” has 4 letters with S repeated twice: \(\dfrac{4!}{2!}\).
Repeated elements: divide by factorials of repeat counts: \(\dfrac{n!}{n_1!\,n_2!\cdots}\).
Bit Strings
Bit strings, combinations, and even/odd patterns
Learning goal: Use combinations to count bit strings with restrictions (like “exactly \(k\) ones” or “an even number of ones”).
Key idea
A bit string of length \(n\) is a sequence of \(0\)s and \(1\)s. To count strings with exactly \(k\) ones, choose which \(k\) positions are ones: \[ \binom{n}{k}. \] To count strings with an even number of ones, you can sum \(\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots\). A key fact (for \(n\ge 1\)) is that exactly half of all \(2^n\) bit strings have an even number of ones, so the count is \(2^{n-1}\).
Worked example
Example: How many bit strings of length 4 have exactly 2 ones?
Choose 2 of the 4 positions to be ones: \[ \binom{4}{2}=6. \] Then the number with an even number of ones is: \[ \binom{4}{0}+\binom{4}{2}+\binom{4}{4}=1+6+1=8. \]
Try it
Try it 1: How many bit strings of length 4 have an even number of ones?
Hint: Half of all \(2^4=16\) bit strings have an even number of ones.
Try it 2: How many bit strings of length 5 have an even number of ones?
Hint: Half of all \(2^5=32\) bit strings have an even number of ones.
Summary
Exactly \(k\) ones in length \(n\): \(\binom{n}{k}\).
Even number of ones (for \(n\ge 1\)): \(2^{n-1}\).
Diagonals & Binomials
Counting diagonals and computing larger binomial coefficients
Learning goal: Use combinations to count pairs and apply formulas efficiently for classic geometry counts.
Key idea
A diagonal connects two non-adjacent vertices of a polygon. To count diagonals in a convex \(n\)-gon:
Choose 2 vertices to form a segment: \(\binom{n}{2}\).
Subtract the \(n\) sides (edges).
So: \[ \text{diagonals}=\binom{n}{2}-n=\frac{n(n-1)}{2}-n=\frac{n(n-3)}{2}. \]
Worked example
Example: How many diagonals does a convex hexagon have?
With \(n=6\): \[ \frac{6(6-3)}{2}=\frac{6\cdot 3}{2}=9. \]
Geometry & graphs: diagonals, edges, and counting pairs.
Worked example: committee + leader
Example: From 5 students, how many ways can you choose a 3-person committee and then select 1 committee leader?
First choose the committee: \(\binom{5}{3}=10\). Then choose the leader from the 3 committee members: \(3\) choices. \[ 10\cdot 3=30. \] This matches the permutation idea \(P(5,3)=60\) divided by \(2!\) (because the two non-leader members are unordered).
Bit strings: exactly \(k\) ones \(\Rightarrow \binom{n}{k}\); even number of ones \(\Rightarrow 2^{n-1}\) for \(n\ge 1\).
Diagonals: \(\dfrac{n(n-3)}{2}\) in a convex \(n\)-gon.
Next step: Close this lesson and try your quiz again. If you miss a question, reopen the book and review the page that matches the counting skill you need.
Practice set
Permutations & Combinations practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
0/10answered
Question 1Not answered
What is \(\binom{4}{1}\)?
Correct answer: C. \(4\)
Explanation: Choosing \(1\) item from \(4\) can be done in \(4\) ways.
Question 2Not answered
How many ways are there to choose \(2\) people from a group of \(7\)?
