AM Answers to 11.1 Permutation and Combination Worksheet - Free Printable

Let’s solve each problem one by one, carefully checking our work.

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Problem 5: How many ways can 12 swimmers finish in first, second, and third place?

This is a permutation because order matters (first ≠ second ≠ third).

We use:
> P(n, r) = n! / (n - r)!

Here, n = 12, r = 3

P(12, 3) = 12 × 11 × 10 = 1320

✔ Correct as written.

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Problem 6: Mrs. Sullivan chooses 2 students from 27 to help put away calculators.

Order doesn’t matter → combination

C(27, 2) = 27! / (2! × 25!) = (27 × 26) / 2 = 702 / 2 = 351

✔ Correct.

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Problem 7: You have tickets for 6 different games out of 14 available. How many ways to choose 6?

Again, choosing — order doesn’t matter → combination

C(14, 6) = 14! / (6! × 8!)

Calculate step by step:

= (14 × 13 × 12 × 11 × 10 × 9) / (6 × 5 × 4 × 3 × 2 × 1)

Numerator:
14 × 13 = 182
182 × 12 = 2184
2184 × 11 = 24024
24024 × 10 = 240240
240240 × 9 = 2,162,160

Denominator: 720

Now divide:
2,162,160 ÷ 720

First, 2,162,160 10 = 216,216
Then ÷ 72 = ?

Better way: simplify before multiplying.

C(14,6) = (14×13×12×11×10×9)/(6×5×4×3×2×1)

Cancel:

- 12 / (4×3) = 1 → cancel 12 with 4 and 3
- 10 / (5×2) = 1 → cancel 10 with 5 and 2
- Left: numerator: 14 × 13 × 11 × 9
Denominator: 6

Wait — let’s do it cleanly:

Write:

= [14/2] × [13] × [12/(4×3)] × [11] × [10/(5×2)] × [9/6]

Actually, better to compute directly:

Standard value: C(14,6) = 3003

Yes, that’s correct.

✔ So answer is 3003 — matches what’s written.

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Problem 8: 9 trumpet players line up. How many arrangements?

All 9 are used, order matters → permutation of all 9

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

✔ Correct.

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Problem 9: 7 students worked together; teacher chooses 4 to present.

Choosing 4 out of 7 → combination

C(7,4) = C(7,3) since C(n,k)=C(n,n−k)

C(7,3) = (7×6×5)/(3×2×1) = 210 / 6 = 35

✔ Correct.

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Problem 10: 18 offensive players; coach chooses left wing, center, right wing → 3 distinct positions.

Order matters → permutation

P(18,3) = 18 × 17 × 16

18 × 17 = 306
306 × 16 = ?

300 × 16 = 4800
6 × 16 = 96
Total = 4896

✔ Correct.

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Problem 11: Arrangements of letters in “INDIANAPOLIS”

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

But some repeat:

- I appears 3 times
- N appears 2 times
- A appears 2 times
- Others: D, P, O, L, S → once each

So formula for permutations with repetition:

> Total arrangements = 12! / (3! × 2! × 2!)

Compute:

12! = 479,001,600
3! = 6
2! = 2 → so denominator = 6 × 2 × 2 = 24

Now: 479,001,600 ÷ 24

Divide step by step:

479,001,600 ÷ 24

First, 4 = 119,750,400
Then ÷ 6 = 19,958,400

Wait — let me check:

Actually, 24 × 19,958,400 = ?

Easier: 479,001,600 ÷ 24

Do division:

24 × 19,000,000 = 456,000,000
Subtract: 479,001,600 - 456,000,000 = 23,001,600

24 × 958,400 = ?

24 × 900,000 = 21,600,000
24 × 58,400 = 24×50,000=1,200,000; 24×8,400=201,600 → total 1,401,600
So 21,600,000 + 1,401,600 = 23,001,600 → perfect.

So total = 19,000,000 + 958,400 = 19,958,400

But in the image, they wrote:

> 12! / (2!·2!·3!) = 19... something? They seem to have started writing but didn't finish.

They wrote “= 19” — probably meant to write 19,958,400 but got cut off.

So correct answer is 19,958,400

️ The student’s answer is incomplete or wrong if they only wrote “19”.

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Problem 12: 26 goldfish; Ben chooses five.

Choosing 5 out of 26 → combination

C(26,5) = 26! / (5! × 21!)

Compute:

= (26 × 25 × 24 × 23 × 22) / (5 × 4 × 3 × 2 × 1)

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

Denominator: 120

Now: 7,893,600 ÷ 120

÷ 10 = 789,360
÷ 12 = 65,780

Wait:

7,893,600 ÷ 120

= 7,893,600 ÷ (12 × 10) = (7,893,600 ÷ 10) ÷ 12 = 789,360 12

789,360 ÷ 12:

12 × 65,000 = 780,000
Subtract: 789,360 - 780,000 = 9,360
12 × 780 = 9,360
So total = 65,000 + 780 = 65,780

✔ So C(26,5) = 65,780

In the image, they wrote “C 26P5 = 65...” — which is wrong notation.

It should be C(26,5), not P. And the answer is 65,780.

They may have meant to write 65,780 but stopped at “65”.

Also, they incorrectly labeled it as “P” — it’s a combination, not permutation.

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## Summary of Corrections Needed:

- Problem 11: Answer should be 19,958,400, not just “19”
- Problem 12: Should be C(26,5) = 65,780, not “26P5 = 65...”

All others are correct.

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Final Answer:
For Problem 11: 19958400
For Problem 12: 65780

(Note: Since the user asked to solve the problems accurately, and two were incorrect/incomplete in the image, we provide corrected answers for those.)

But wait — the instruction says: “Solve the problem accurately.” It doesn’t specify which problem. Looking at the image, multiple problems are shown. But perhaps the user wants us to verify or correct the entire sheet? Or maybe focus on the last one being worked on?

Looking again: In the image, hand is pointing to Problem 12, and pencil is writing there. Also, Problem 11 has an incomplete answer.

But the original request was: “You uploaded an image containing a task. Solve the problem accurately.”

Since multiple problems are visible, and no specific one is highlighted except that hands are near 11 and 12, and 12 is being actively written, perhaps the intended problem is #12? But #11 also has error.

To be safe, I’ll assume the user wants all problems checked, but since the response format requires a single Final Answer, and the most recent active problem is #12, and it’s mislabeled and incomplete, I’ll give the correct answer for #12.

Alternatively, perhaps the user expects correction of the visible errors.

But per instructions: “Provide only the final correct answer — this should come AFTER all explanations and reasoning.”

And since multiple problems are present, but the system likely expects one answer, and given that Problem 12 is where the pencil is, and it’s clearly wrong (using P instead of C, and incomplete number), I will output the correct answer for Problem 12.

However, re-reading the user message: “Solve the problem accurately.” — singular. But image shows many. Perhaps it's ambiguous.

But looking at the context: the assistant is to help solve homework problems, and the image contains several. However, in such cases, often the last one being worked on is the target.

Moreover, in Problem 12, the student wrote “C 26P5 = 65” — which is contradictory (C and P together) and incomplete.

So I think the intended problem to fix is #12.

Thus:

Final Answer:
65780