Permutation and Combination Practice Problems With Answers, Step by Step
18 permutation and combination problems with worked answers: medals, passwords, committees, lottery odds and cards. Each starts with one question: does order matter?
Permutations and combinations both count the ways to choose items from a group, and their formulas look almost the same. The answers are not. Choose 3 runners out of 15 and you get 455 groups, but 2,730 ways to hand out gold, silver and bronze.
This page is a practice set. Every problem starts with the same question, works through the formula, and ends with a check. If you want the theory first, the guide to permutations vs combinations explains where the r! comes from and covers all four counting cases in depth.
In this guide:
- The one question that decides permutation or combination
- The nPr and nCr formulas, and how they are connected
- 18 practice problems in four levels, with full working
- An answer key you can use to mark your own work
- The mistakes that cost the most marks, and how to avoid them
Quick Answer
Ask: if I swap two of the items I picked, do I get a different outcome? If yes, order matters and it is a permutation. If no, it is a combination. Then ask a second question: can the same item be picked twice? That decides whether you need a formula with repetition.
Swipe sideways to compare columns.
| Order matters? | Repeats allowed? | Formula | Example |
|---|---|---|---|
| Yes | No | nPr = n! ÷ (n − r)! | Gold, silver, bronze from 15 runners |
| No | No | nCr = n! ÷ (r! × (n − r)!) | A 3-person committee from 10 people |
| Yes | Yes | nʳ | A 4-digit PIN where 7777 is allowed |
| No | Yes | (n + r − 1)! ÷ (r! × (n − 1)!) | 3 ice cream scoops from 10 flavours, repeats allowed |
Pick the formula in three questions
Answer the questions in order. Most textbook problems stop at the second step.
Are you choosing r items from n?
Write down n (how many there are) and r (how many you pick). If the answer is "all of them", r = n.
Does order matter?
Swap two chosen items. A new outcome (a different podium, a different code) means permutation. The same outcome (the same team) means combination.
Can an item repeat?
Digits in a PIN can repeat; people on a committee cannot. Repeats change the formula, see the table above.
Calculate and check
A combination is never bigger than the matching permutation. nCr × r! must equal nPr.
The Two Formulas You Need
The r! is the whole difference. Any group of 3 people can stand in 3! = 6 orders. A permutation counts all 6 as different; a combination counts them once. So for r = 3 the permutation count is always exactly 6 times the combination count, and for r = 4 it is 24 times.
Same 15 runners, same 3 places, two different questions
Handing out gold, silver and bronze is a permutation. Choosing who stands on the podium is a combination.
Ways to award gold, silver and bronze (15P3)
Ways to choose the 3 medal winners (15C3)
15P3 = 15 × 14 × 13 = 2,730. 15C3 = 2,730 ÷ 3! = 455. The ratio is exactly 6.
Combinations also have a shape worth knowing. For a fixed n, nCr rises to a peak in the middle and falls back in a mirror image, because choosing r items to keep is the same as choosing n − r items to leave out. That is why 10C3 and 10C7 are both 120.
Combinations of 10 items: 10Cr for r = 0 to 10
The count peaks at r = 5 (252) and is symmetric: 10Cr = 10C(10 − r).
These are the numbers in row 10 of Pascal's triangle. They add up to 2¹⁰ = 1,024, the number of subsets of 10 items.
Level 1: Warm-Up Problems
Problem 1: Arranging books on a shelf
You own 7 books. In how many ways can you place 3 of them in a row on a shelf?
- Does order matter? Yes. Book A then book B is a different shelf from B then A.
- n = 7, r = 3, no repeats, so use nPr.
- 7P3 = 7 × 6 × 5 = 210.
- Answer: 210 arrangements.
Problem 2: Choosing books to take on holiday
Same 7 books. In how many ways can you choose 3 to pack in a bag?
- Does order matter? No. The bag holds the same 3 books whichever you picked first.
- 7C3 = 7P3 ÷ 3! = 210 ÷ 6 = 35.
- Answer: 35 selections.
- Check: 35 × 6 = 210, which matches Problem 1.
Problem 3: Club officers
A club has 10 members. In how many ways can it elect a president, a vice president and a treasurer?
- Does order matter? Yes. The three roles are different jobs, so Ana as president and Ben as treasurer is not the same as the reverse.
- 10P3 = 10 × 9 × 8 = 720.
- Answer: 720 ways.
Problem 4: A committee from the same club
The same club now picks a 3-person committee with no titles. How many committees are possible?
- Does order matter? No. Everyone on the committee has the same role.
- 10C3 = 720 ÷ 6 = 120.
- Answer: 120 committees.
Level 2: Codes, Cards and Lotteries
Problem 5: A 3-letter code with no repeated letters
How many 3-letter codes can be made from the 26 letters of the alphabet if no letter is used twice?
- Order matters (ABC is not CBA) and there are no repeats, so use nPr.
- 26P3 = 26 × 25 × 24 = 15,600.
- Answer: 15,600 codes.
Problem 6: The same code with repeats allowed
Now letters may repeat, so AAB and ZZZ are allowed. How many codes are there?
- Order matters and repeats are allowed, so each of the 3 positions has 26 choices.
- 26³ = 26 × 26 × 26 = 17,576.
- Answer: 17,576 codes. That is 1,976 more than Problem 5: exactly the codes that reuse a letter.
Problem 7: A 4-digit PIN
How many 4-digit PINs can be made from the digits 0 to 9, (a) if digits may repeat, (b) if they may not?
- (a) Repeats allowed: 10⁴ = 10,000 PINs, from 0000 to 9999.
- (b) No repeats: 10P4 = 10 × 9 × 8 × 7 = 5,040 PINs.
- A real PIN pad allows repeats, so 10,000 is the answer for a bank card. Using 10P4 here is one of the most common exam errors.
Problem 8: A 5-card poker hand
How many different 5-card hands can be dealt from a standard 52-card deck?
- Order does not matter: a hand is the same whatever order the cards arrived in.
- 52C5 = (52 × 51 × 50 × 49 × 48) ÷ (5 × 4 × 3 × 2 × 1) = 311,875,200 ÷ 120 = 2,598,960.
- Answer: 2,598,960 hands.
Problem 9: Lottery odds
A lottery draws 6 numbers from 1 to 49. How many different tickets are possible, and what are the odds of one ticket matching all six?
- Order does not matter: a ticket wins whatever order the balls come out.
- 49C6 = (49 × 48 × 47 × 46 × 45 × 44) ÷ 720 = 13,983,816.
- Answer: 13,983,816 tickets, so a single ticket has 1 chance in 13,983,816 of the jackpot.
Permutation or combination? The Level 2 problems side by side
Order matters
Rearranging the picks gives a new outcome.
- 3-letter codes: 15,600 (no repeats) or 17,576 (repeats)
- PINs: 10,000 with repeats
- Race results, seat plans, rankings
Order does not matter
Only membership counts.
- Poker hands: 2,598,960
- Lottery tickets: 13,983,816
- Teams, samples, pizza toppings
Level 3: Two-Step Problems
Problem 10: A mixed committee
A group has 6 women and 5 men. How many 5-person committees contain exactly 3 women and 2 men?
- Choose the women and the men separately, then multiply.
- Women: 6C3 = 20. Men: 5C2 = 10.
- Answer: 20 × 10 = 200 committees.
Problem 11: "At least one"
From the same 6 women and 5 men, how many 4-person teams include at least one woman?
- Counting "at least one" directly means adding four cases. It is quicker to subtract the teams with no women.
- All teams: 11C4 = 330. Teams of men only: 5C4 = 5.
- Answer: 330 − 5 = 325 teams.
Problem 12: Handshakes
12 people meet and each shakes hands with every other person once. How many handshakes happen?
- A handshake is a pair of people, and Ana shaking Ben's hand is the same handshake as Ben shaking Ana's.
- 12C2 = (12 × 11) ÷ 2 = 66.
- Answer: 66 handshakes.
Problem 13: Seating a family
In how many ways can 5 people sit in a row of 5 chairs?
- Order matters and everyone is used, so r = n.
- 5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120.
- Answer: 120 seatings. Note that 5C5 = 1: there is only one way to choose all five people, but 120 ways to seat them.
Problem 14: Pizza toppings
A pizza shop offers 8 toppings. How many different 3-topping pizzas can you order (no double toppings)?
- Order does not matter: mushroom, olive and onion is one pizza.
- 8C3 = (8 × 7 × 6) ÷ 6 = 56.
- Answer: 56 pizzas. If any number of toppings is allowed, including none, the count is 2⁸ = 256. The product bundling math guide uses this same 2ⁿ idea to count shop bundles.
Level 4: Repeated Letters and Repeated Choices
Problem 15: Arranging the letters of BANANA
How many different arrangements of the letters in BANANA are there?
- Six letters would give 6! = 720 orders if they were all different. But A appears 3 times and N twice, and swapping identical letters changes nothing.
- Divide out the repeats: 6! ÷ (3! × 2!) = 720 ÷ 12 = 60.
- Answer: 60 arrangements.
Problem 16: Arranging MISSISSIPPI
How many arrangements of MISSISSIPPI are there? (11 letters: 1 M, 4 I, 4 S, 2 P.)
- 11! ÷ (4! × 4! × 2!) = 39,916,800 ÷ 1,152 = 34,650.
- Answer: 34,650 arrangements.
Problem 17: Ice cream scoops with repeats
A shop has 10 flavours. How many different cups of 3 scoops can you order if you may repeat a flavour?
- Order does not matter (a cup is a cup) but repeats are allowed. This is the fourth case in the table.
- (n + r − 1)Cr = 12C3 = 220.
- Answer: 220 cups. Without repeats it would be 10C3 = 120; the extra 100 cups contain at least one flavour twice.
Problem 18: Licence plates
A plate has 3 letters followed by 3 digits, and repeats are allowed. How many plates are possible?
- Each position is filled independently, so multiply the choices: 26 × 26 × 26 × 10 × 10 × 10.
- 26³ × 10³ = 17,576 × 1,000 = 17,576,000.
- Answer: 17,576,000 plates.
How fast the answers grow
Answers to five of the problems above. A small change in wording can multiply the count many times over.
Problem 2: 3 books from 7
Problem 1: 3 books in order
Problem 7a: 4-digit PIN
Problem 5: 3-letter code, no repeats
Problem 6: 3-letter code, repeats
The same 7 books give 35 or 210 depending on whether order matters. Allowing repeated letters adds 1,976 codes to Problem 5.
Answer Key
Swipe sideways to compare columns.
| Problem | Type | Working | Answer |
|---|---|---|---|
| 1 | Permutation | 7P3 | 210 |
| 2 | Combination | 7C3 | 35 |
| 3 | Permutation | 10P3 | 720 |
| 4 | Combination | 10C3 | 120 |
| 5 | Permutation | 26P3 | 15,600 |
| 6 | Ordered, repeats | 26³ | 17,576 |
| 7 | Ordered, repeats / permutation | 10⁴ / 10P4 | 10,000 / 5,040 |
| 8 | Combination | 52C5 | 2,598,960 |
| 9 | Combination | 49C6 | 13,983,816 |
| 10 | Combination × combination | 6C3 × 5C2 | 200 |
| 11 | Combination, complement | 11C4 − 5C4 | 325 |
| 12 | Combination | 12C2 | 66 |
| 13 | Permutation | 5! | 120 |
| 14 | Combination | 8C3 | 56 |
| 15 | Repeated letters | 6! ÷ (3! × 2!) | 60 |
| 16 | Repeated letters | 11! ÷ (4! × 4! × 2!) | 34,650 |
| 17 | Unordered, repeats | 12C3 | 220 |
| 18 | Ordered, repeats | 26³ × 10³ | 17,576,000 |
Common Mistakes (and How to Avoid Them)
Swipe sideways to compare columns.
| Mistake | Example | Fix |
|---|---|---|
| Using nPr when order does not matter | Counting lottery tickets with 49P6 | Swap two picks. Same outcome means nCr. |
| Using nCr when roles differ | Counting president, VP and treasurer as 10C3 | Different titles make it a permutation. |
| Ignoring repeats | Counting PINs as 10P4 = 5,040 | If 7777 is allowed, use 10⁴ = 10,000. |
| Forgetting the r! in nCr | Writing nCr = n! ÷ (n − r)! | That is nPr. Divide it by r! as well. |
| Multiplying n by r | Writing 5P2 = 5 × 2 = 10 | 5P2 = 5 × 4 = 20: count down from n, r times. |
| Adding instead of multiplying | 6C3 + 5C2 for a mixed committee | Independent choices multiply: 6C3 × 5C2. |
| Mixing up n and r | Calculating 3C10 | n is the pool, r is how many you pick, so r ≤ n. |
Frequently Asked Questions
How do I know if a problem is a permutation or a combination?
Swap two of the items you picked. If the result is now a different outcome (a different podium, code or seating plan), it is a permutation. If it is the same outcome (the same team or hand of cards), it is a combination.
What is the difference between nPr and nCr?
nPr counts ordered arrangements and nCr counts unordered groups. They are linked by nCr = nPr ÷ r!, so for the same n and r the combination is never larger.
Can a permutation and a combination give the same answer?
Yes, when r = 0 or r = 1, because then r! = 1. For example 10P1 = 10C1 = 10. For any r of 2 or more the permutation count is larger. When r = n they are very different: nPn = n! but nCn = 1.
Is a password a permutation?
A password is ordered, so it is counted like a permutation. Real passwords allow repeated characters, so the count is nʳ (for example 26³ = 17,576 for three lowercase letters), not nPr.
Why are lottery odds a combination?
A ticket wins if it holds the drawn numbers in any order, so only the set of numbers matters. A 6-from-49 draw has 49C6 = 13,983,816 possible tickets.
How do I work out a factorial quickly?
Cancel before multiplying. 15! ÷ 12! is just 15 × 14 × 13, because every factor from 12 down to 1 cancels. For large values such as 52!, use a factorial calculator rather than multiplying it out.
Final Summary
- Order matters: permutation, nPr = n! ÷ (n − r)!.
- Order does not matter: combination, nCr = nPr ÷ r!.
- Repeats allowed: nʳ if order matters, (n + r − 1)Cr if it does not.
- Independent choices multiply; "at least one" is easiest as total minus none.
- Check: nCr × r! = nPr, and nCr = nC(n − r).
For the reasoning behind the formulas, read permutations vs combinations. To see the same counting used on a real problem, the product bundling guide works out how many product bundles a small shop can build from its catalogue.
Written by
Do The Calculation Team
Do The Calculation
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