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Permutation worksheet

Grades 11–12 · CCSS HSS.CP.B.9

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Repetition: With repetition
- Number of items to choose from (n): 8
- Number of positions to fill (r): 3
Find: Permutations with repetition
Answer
2.
Problem 2.
- Repetition: With repetition
- Number of items to choose from (n): 7
- Number of positions to fill (r): 2
Find: Permutations with repetition
Answer
3.
Problem 3.
- Repetition: Without repetition
- Number of items to choose from (n): 7
- Number of positions to fill (r): 3
Find: Permutations, P(n, r)
Answer
4.
Problem 4.
- Repetition: Without repetition
- Number of items to choose from (n): 9
- Number of positions to fill (r): 4
Find: Permutations, P(n, r)
Answer
5.
Problem 5.
- Repetition: Without repetition
- Number of items to choose from (n): 6
- Number of positions to fill (r): 4
Find: Permutations, P(n, r)
Answer
6.
Problem 6.
- Repetition: Without repetition
- Number of items to choose from (n): 11
- Number of positions to fill (r): 5
Find: Permutations, P(n, r)
Answer
7.
Problem 7.
- Repetition: Without repetition
- Number of items to choose from (n): 10
- Number of positions to fill (r): 3
Find: Permutations, P(n, r)
Answer
8.
Problem 8.
- Repetition: With repetition
- Number of items to choose from (n): 8
- Number of positions to fill (r): 2
Find: Permutations with repetition
Answer
9.
Problem 9.
- Repetition: Without repetition
- Number of items to choose from (n): 8
- Number of positions to fill (r): 2
Find: Permutations, P(n, r)
Answer
10.
Problem 10.
- Repetition: With repetition
- Number of items to choose from (n): 7
- Number of positions to fill (r): 3
Find: Permutations with repetition
Answer

Answer key: Permutation worksheet

1. Permutations with repetition: 512
1. Any of the n items can fill each of the r positions, so multiply n by itself r times
  $N$ $= n^{r}$
2. Put in your numbers: n = 8 items and r = 3 positions
  $8^{3}$ $= 512$
2. Permutations with repetition: 49
1. Any of the n items can fill each of the r positions, so multiply n by itself r times
  $N$ $= n^{r}$
2. Put in your numbers: n = 7 items and r = 2 positions
  $7^{2}$ $= 49$
3. Permutations, P(n, r): 210
1. Order matters and no item is used twice, so use the permutations formula (! means multiply by every whole number below, down to 1)
  $P (n, r)$ $= \frac{n!}{(n - r)!}$
2. Put in your numbers: n = 7 items and r = 3 positions
  $P (7, 3)$ $= \frac{7!}{(7 - 3)!}$ $= \frac{7!}{4!}$
3. Cancel 4! from the top and bottom, since it appears in both. That leaves the top 3 factors of 7! multiplied together
  $7 \times 6 \times 5$ $= 210$
4. Permutations, P(n, r): 3,024
1. Order matters and no item is used twice, so use the permutations formula (! means multiply by every whole number below, down to 1)
  $P (n, r)$ $= \frac{n!}{(n - r)!}$
2. Put in your numbers: n = 9 items and r = 4 positions
  $P (9, 4)$ $= \frac{9!}{(9 - 4)!}$ $= \frac{9!}{5!}$
3. Cancel 5! from the top and bottom, since it appears in both. That leaves the top 4 factors of 9! multiplied together
  $9 \times 8 \times 7 \times 6$ $= 3,024$
5. Permutations, P(n, r): 360
1. Order matters and no item is used twice, so use the permutations formula (! means multiply by every whole number below, down to 1)
  $P (n, r)$ $= \frac{n!}{(n - r)!}$
2. Put in your numbers: n = 6 items and r = 4 positions
  $P (6, 4)$ $= \frac{6!}{(6 - 4)!}$ $= \frac{6!}{2!}$
3. Cancel 2! from the top and bottom, since it appears in both. That leaves the top 4 factors of 6! multiplied together
  $6 \times 5 \times 4 \times 3$ $= 360$
6. Permutations, P(n, r): 55,440
1. Order matters and no item is used twice, so use the permutations formula (! means multiply by every whole number below, down to 1)
  $P (n, r)$ $= \frac{n!}{(n - r)!}$
2. Put in your numbers: n = 11 items and r = 5 positions
  $P (11, 5)$ $= \frac{11!}{(11 - 5)!}$ $= \frac{11!}{6!}$
3. Cancel 6! from the top and bottom, since it appears in both. That leaves the top 5 factors of 11! multiplied together
  $11 \times 10 \times 9 \times 8 \times 7$ $= 55,440$
7. Permutations, P(n, r): 720
1. Order matters and no item is used twice, so use the permutations formula (! means multiply by every whole number below, down to 1)
  $P (n, r)$ $= \frac{n!}{(n - r)!}$
2. Put in your numbers: n = 10 items and r = 3 positions
  $P (10, 3)$ $= \frac{10!}{(10 - 3)!}$ $= \frac{10!}{7!}$
3. Cancel 7! from the top and bottom, since it appears in both. That leaves the top 3 factors of 10! multiplied together
  $10 \times 9 \times 8$ $= 720$
8. Permutations with repetition: 64
1. Any of the n items can fill each of the r positions, so multiply n by itself r times
  $N$ $= n^{r}$
2. Put in your numbers: n = 8 items and r = 2 positions
  $8^{2}$ $= 64$
9. Permutations, P(n, r): 56
1. Order matters and no item is used twice, so use the permutations formula (! means multiply by every whole number below, down to 1)
  $P (n, r)$ $= \frac{n!}{(n - r)!}$
2. Put in your numbers: n = 8 items and r = 2 positions
  $P (8, 2)$ $= \frac{8!}{(8 - 2)!}$ $= \frac{8!}{6!}$
3. Cancel 6! from the top and bottom, since it appears in both. That leaves the top 2 factors of 8! multiplied together
  $8 \times 7$ $= 56$
10. Permutations with repetition: 343
1. Any of the n items can fill each of the r positions, so multiply n by itself r times
  $N$ $= n^{r}$
2. Put in your numbers: n = 7 items and r = 3 positions
  $7^{3}$ $= 343$