How it works
If you arrange n different objects in a row, there are n choices for the first spot, n − 1 for the next, and so on, giving
n! = n × (n − 1) × … × 2 × 1
If you only fill k spots (gold, silver and bronze from 8 runners), stop after k factors: P(n, k) = n! ÷ (n − k)!. So 8 runners give 8 × 7 × 6 = 336 podiums.
Three tricks contests use
Glue: if certain people must stand together, treat them as one block, arrange the blocks, then multiply by the ways to arrange inside each block. Complement: "not next to each other" is easiest as (all arrangements) − (arrangements where they are together). Repeats: if letters repeat, divide by the factorial of each repeat count, so AABB has 4! ÷ (2! × 2!) = 6 arrangements.
Permutations practice problems with solutions
These problems come from Mathyessy's question bank. Try each one before opening the solution.
How many ways to award gold and silver among 5 runners?
Show answer and solution
Answer: 20
Order matters: 5 choices for gold, then 4 for silver, so 5 × 4 = 20.
How many 3-digit PINs with no repeated digits? The first digit may be zero.
Show answer and solution
Answer: 720
10 choices for the first digit, 9 for the second and 8 for the third: 10 × 9 × 8 = 720.
Six books are arranged in a row. Two specific books must NOT be next to each other. How many arrangements are possible?
Show answer and solution
Answer: 480
All arrangements: 6! = 720. Arrangements with the two books together: glue them into one block, giving 5! × 2 = 240. So 720 − 240 = 480 keep them apart.
How many distinct arrangements of 4 letters with one letter 2 times and another 2 times?
Show answer and solution
Answer: 6
Arrangements of AABB: 4! ÷ (2! × 2!) = 24 ÷ 4 = 6. (AABB, ABAB, ABBA, BAAB, BABA, BBAA.)
Keep going
Related topics
Related practice sets in the app
All Counting & Probability topics · Contest math practice by topic · MATHCOUNTS practice · AMC 8 practice