Solving combinatorics problems on a math worksheet with handwritten solutions.

It looks like you've uploaded a photo of a math worksheet with several combinatorics problems (permutations and combinations), and some answers have already been filled in. Let's go through each problem one by one, verify the solutions, and explain the reasoning behind them.

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Problem 5:

> How many ways can 12 swimmers finish in first, second, and third place?

This is about ranking — order matters. So we use permutations.

We want to choose 3 swimmers out of 12 and arrange them in order.

$$
P(12, 3) = 12 \times 11 \times 10 = 1320
$$

✔ The answer written is 1320 → Correct

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Problem 6:

> How many ways can Mrs. Sullivan choose two students from 27 to help put away calculators at the end of class?

Order does not matter here (just choosing two students). Use combinations.

$$
C(27, 2) = \frac{27 \times 26}{2} = 351
$$

✔ The answer written is 351 → Correct

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Problem 7:

> You have enough tickets to play 6 different games at the amusement park. There are 14 games. How many ways can you choose 6?

Again, order doesn’t matter — just selecting 6 out of 14.

$$
C(14, 6) = \frac{14!}{6!(14-6)!} = 3003
$$

✔ The answer written is 3003 → Correct

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Problem 8:

> How many different ways can 9 trumpet players in the marching band line up?

All 9 players are being arranged — full permutation.

$$
P(9, 9) = 9! = 362,880
$$

✔ The answer written is 362,880 → Correct

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Problem 9:

> 18 offensive players on the hockey team. How many ways can the coach choose a left wing, center, and right wing to start the game?

Now, positions are distinct (left wing, center, right wing) — so order matters.

We’re choosing 3 specific roles from 18 players, and assigning them to positions.

So this is a permutation:

$$
P(18, 3) = 18 \times 17 \times 16 = 4896
$$

✔ The answer written is 4896 → Correct

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Problem 10:

> How many different 12-letter arrangements can be made using the letters in the word INDIANAPOLIS?

This is a permutation of multiset — repeated letters.

First, count total letters: 12
Letters: I, N, D, I, A, N, A, P, O, L, I, S

Let’s count frequencies:
- I: 3 times
- N: 2 times
- A: 2 times
- D, P, O, L, S: 1 each

Total arrangements:

$$
\frac{12!}{3! \cdot 2! \cdot 2! \cdot 1! \cdot 1! \cdot 1! \cdot 1! \cdot 1!} = \frac{479001600}{6 \cdot 2 \cdot 2} = \frac{479001600}{24} = 19,958,400
$$

But the student wrote: $ \frac{12!}{(2!)(2!)(3!)} = 19 $ → That’s incorrect.

Wait — that’s not possible. The answer should be 19,958,400, not 19.

✘ The answer written is 19 → Incorrect

The formula is correct, but the result is wrong.

Actually, let's check what they did:

They wrote:
$ \frac{12!}{(2!)(2!)(3!)} $

That’s correct for the denominator: 2! for A, 2! for N, 3! for I.

But $ \frac{12!}{2! \cdot 2! \cdot 3!} = \frac{479001600}{6 \cdot 2 \cdot 2} = \frac{479001600}{24} = 19,958,400 $

But they wrote 19 — probably a mistake in calculation or notation.

Maybe they meant to write the number of distinct permutations, but accidentally wrote 19 instead of 19,958,400.

So: ✘ Incorrect answer.

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Problem 11:

> There are 26 gold fish in the tank at the store. How many ways can Ben choose five?

Choosing 5 out of 26, order doesn't matter → combination.

$$
C(26, 5) = \frac{26!}{5!(26-5)!} = \frac{26 \times 25 \times 24 \times 23 \times 22}{5 \times 4 \times 3 \times 2 \times 1}
$$

Calculate step by step:

Numerator:
26 × 25 = 650
650 × 24 = 15,600
15,600 × 23 = 358,800
358,800 × 22 = 7,893,600

Denominator: 120

Now divide:
7,893,600 ÷ 120 = 65,780

✔ The answer written is 65,780 → Correct

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Summary of Corrections:

| Problem | Correct? | Explanation |
|--------|---------|-----------|
| 5 | ✔ Yes | $ P(12,3) = 1320 $ |
| 6 | ✔ Yes | $ C(27,2) = 351 $ |
| 7 | ✔ Yes | $ C(14,6) = 3003 $ |
| 8 | ✔ Yes | $ 9! = 362,880 $ |
| 9 | ✔ Yes | $ P(18,3) = 4896 $ |
| 10 | ✘ No | Should be $ \frac{12!}{3! \cdot 2! \cdot 2!} = 19,958,400 $, not 19 |
| 11 | ✔ Yes | $ C(26,5) = 65,780 $ |

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Final Answer:

Only Problem 10 has an incorrect answer.

✔ Correct answer for Problem 10:
$$
\frac{12!}{3! \cdot 2! \cdot 2!} = 19,958,400
$$

Let me know if you'd like a detailed breakdown of how to compute it step by step!