Permutations and Combinations worksheet - Free Printable

Let's solve the Permutations and Combinations worksheet step by step.

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I. Evaluate

We will use the formulas:

- Permutation:
$$
_nP_r = \frac{n!}{(n - r)!}
$$

- Combination:
$$
_nC_r = \frac{n!}{r!(n - r)!}
$$

Also recall:
- $ 0! = 1 $
- $ _nP_0 = 1 $
- $ _nC_0 = 1 $
- $ _nP_n = n! $
- $ _nC_n = 1 $

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#### 1. $ _{12}P_3 $

$$
= \frac{12!}{(12 - 3)!} = \frac{12!}{9!} = 12 \times 11 \times 10 = 1320
$$

✔ Answer: 1320

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#### 2. $ _{12}C_3 $

$$
= \frac{12!}{3!(12 - 3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220
$$

✔ Answer: 220

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#### 3. $ _{20}P_2 $

$$
= \frac{20!}{(20 - 2)!} = \frac{20!}{18!} = 20 \times 19 = 380
$$

✔ Answer: 380

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#### 4. $ _{20}C_2 $

$$
= \frac{20!}{2!(20 - 2)!} = \frac{20 \times 19}{2 \times 1} = 190
$$

✔ Answer: 190

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#### 5. $ _9P_1 $

$$
= \frac{9!}{(9 - 1)!} = \frac{9!}{8!} = 9
$$

✔ Answer: 9

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#### 6. $ _9C_1 $

$$
= \frac{9!}{1!(9 - 1)!} = \frac{9}{1} = 9
$$

✔ Answer: 9

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#### 7. $ _9P_9 $

$$
= \frac{9!}{(9 - 9)!} = \frac{9!}{0!} = 9! = 362880
$$

✔ Answer: 362880

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#### 8. $ _9C_9 $

$$
= \frac{9!}{9!(9 - 9)!} = \frac{9!}{9! \cdot 0!} = 1
$$

✔ Answer: 1

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#### 9. $ _9P_0 $

$$
= \frac{9!}{(9 - 0)!} = \frac{9!}{9!} = 1
$$

✔ Answer: 1

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#### 10. $ _9C_0 $

$$
= \frac{9!}{0!(9 - 0)!} = \frac{9!}{1 \cdot 9!} = 1
$$

✔ Answer: 1

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✔ Summary of Part I:

| Problem | Answer |
|--------|--------|
| 1. $ _{12}P_3 $ | 1320 |
| 2. $ _{12}C_3 $ | 220 |
| 3. $ _{20}P_2 $ | 380 |
| 4. $ _{20}C_2 $ | 190 |
| 5. $ _9P_1 $ | 9 |
| 6. $ _9C_1 $ | 9 |
| 7. $ _9P_9 $ | 362880 |
| 8. $ _9C_9 $ | 1 |
| 9. $ _9P_0 $ | 1 |
| 10. $ _9C_0 $ | 1 |

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II. Set up a notation in the form of $ _nP_r $ or $ _nC_r $, then find the number of possibilities

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#### 1. Selecting which ten players will be in the relay game on a 15-person team.

- We are choosing 10 out of 15, and order does not matter (just selecting the team).
- So it’s a combination.

Set up: $ _{15}C_{10} $

Now calculate:

$$
_{15}C_{10} = \frac{15!}{10! \cdot 5!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = \frac{360360}{120} = 3003
$$

✔ Answer: 3003

> Therefore, there are 3003 different ways to select 10 players from 15.

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#### 2. Kelly is setting a four-digit lock using digits 2, 4, 1, 3, and 8. He wants to use all five digits? Wait — he has five digits but only a four-digit lock**.

But the problem says: *"He is considering different ways to use the digits 2, 4, 1, 3, and 8."* — so he's choosing 4 digits out of 5, and arranging them.

Wait — actually, the wording says "use the digits 2, 4, 1, 3, and 8" — that’s five digits, but the lock is four-digit.

So likely: He wants to use 4 of these 5 digits, and arrange them in order.

But let’s re-read: “different ways to use the digits 2, 4, 1, 3, and 8” — possibly implying using all five digits, but that doesn’t fit a four-digit lock.

Wait — maybe he’s using all four digits from the set, but the set has five digits: 2, 4, 1, 3, 8.

Ah! Probably: He will choose 4 digits from the 5 available, and arrange them in the lock.

But the question says: “use the digits 2, 4, 1, 3, and 8” — meaning these are the digits available, and he’s forming a four-digit code using some or all?

But if he uses only 4 digits, and they are distinct, and order matters, and no repetition.

So: Choose 4 digits from 5, and permute them.

So:

Set up: $ _5P_4 $

$$
= \frac{5!}{(5 - 4)!} = \frac{120}{1} = 120
$$

✔ Answer: 120

> Therefore, there are 120 different ways to use 4 digits from the 5 given in a four-digit lock.

Alternatively, if he must use all five digits, that wouldn't work for a four-digit lock. So the only logical interpretation is: selecting 4 distinct digits from 5 and arranging them.

So: $ _5P_4 = 120 $

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#### 3. There are 25 applicants for five IT expert positions.

- We are selecting 5 people from 25.
- Since it's positions, we assume order does not matter (unless specified otherwise), so it's a combination.

Set up: $ _{25}C_5 $

$$
= \frac{25!}{5! \cdot 20!} = \frac{25 \times 24 \times 23 \times 22 \times 21}{5 \times 4 \times 3 \times 2 \times 1}
$$

Calculate numerator:
$$
25 × 24 = 600 \\
600 × 23 = 13,800 \\
13,800 × 22 = 303,600 \\
303,600 × 21 = 6,375,600
$$

Denominator: $ 5! = 120 $

Now divide:
$$
\frac{6,375,600}{120} = 53,130
$$

✔ Answer: 53,130

> Therefore, there are 53,130 different ways to select 5 from 25 applicants.

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#### 4. A team of 13 volleyball players needs to choose a captain and co-captain.

- Two distinct roles: captain and co-captain → order matters
- So, we’re selecting 2 people from 13, where order matters → permutation

Set up: $ _{13}P_2 $

$$
= \frac{13!}{(13 - 2)!} = 13 × 12 = 156
$$

✔ Answer: 156

> Therefore, there are 156 different ways to choose a captain and co-captain.

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✔ Summary of Part II:

| Problem | Set Up | Answer |
|--------|--------|--------|
| 1. Select 10 players from 15 | $ _{15}C_{10} $ | 3003 |
| 2. Four-digit lock with 5 digits (choose 4) | $ _5P_4 $ | 120 |
| 3. Choose 5 from 25 applicants | $ _{25}C_5 $ | 53,130 |
| 4. Captain and co-captain from 13 players | $ _{13}P_2 $ | 156 |

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✔ Final Answers:

#### Part I: Evaluate

1. $ _{12}P_3 = 1320 $
2. $ _{12}C_3 = 220 $
3. $ _{20}P_2 = 380 $
4. $ _{20}C_2 = 190 $
5. $ _9P_1 = 9 $
6. $ _9C_1 = 9 $
7. $ _9P_9 = 362880 $
8. $ _9C_9 = 1 $
9. $ _9P_0 = 1 $
10. $ _9C_0 = 1 $

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#### Part II: Set up and solve

1. Set up: $ _{15}C_{10} $ → 3003
2. Set up: $ _5P_4 $ → 120
3. Set up: $ _{25}C_5 $ → 53,130
4. Set up: $ _{13}P_2 $ → 156

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