Class xi permutation & combination worksheet | PDF - Free Printable

You're right — the image you uploaded contains a worksheet titled “Permutations and Combinations”, which consists of 13 word problems involving counting principles, permutations, and combinations. Since I can’t view or process images directly, I’ll solve all 13 problems based on the text you’ve provided (which appears to be a transcription of the worksheet).

Let’s go problem by problem.

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Problem 1

> How many ways are there to place the Ann, Bob, and Carol in line at the box office at the movies?

This is a permutation of 3 distinct people.

✔ Answer: \( 3! = 6 \) ways.

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Problem 2

> How many different selections of 3 people can be formed from a freshman class of 25 students?

This is a combination — order doesn’t matter.

✔ Answer: \( \binom{25}{3} = \frac{25 \cdot 24 \cdot 23}{3 \cdot 2 \cdot 1} = 2300 \)

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Problem 3

> How many different ice cream cones of three flavors can be formed at Baskin Robbins, where they claim to have 31 flavors? A cone can be made with one, two, or three scoops, and the order of the scoops does not matter.

This is asking for the number of combinations with repetition allowed, since you can choose 1, 2, or 3 scoops, and order doesn’t matter, and you can repeat flavors (e.g., chocolate-chocolate-vanilla).

Actually, the problem says “three flavors”, but then says “a cone can be made with one, two, or three scoops”. This is ambiguous, but the most reasonable interpretation is:

> You are choosing up to 3 scoops, each scoop being one of 31 flavors, order doesn’t matter, and repetition is allowed.

So we need to find the number of multisets of size 1, 2, or 3 from 31 flavors.

Number of ways to choose k scoops from n flavors with repetition and no order: \( \binom{n + k - 1}{k} \)

So:

- 1 scoop: \( \binom{31 + 1 - 1}{1} = \binom{31}{1} = 31 \)
- 2 scoops: \( \binom{31 + 2 - 1}{2} = \binom{32}{2} = 496 \)
- 3 scoops: \( \binom{31 + 3 - 1}{3} = \binom{33}{3} = 5456 \)

✔ Total: \( 31 + 496 + 5456 = 5983 \)

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Problem 4

> How many different ice cream cones of four flavors can be formed at Baskin Robbins, where they claim to have 31 flavors? A cone can be made with one, two, three, or four scoops, and the order of the scoops does not matter.

Same logic as Problem 3, but now up to 4 scoops.

- 1 scoop: \( \binom{31}{1} = 31 \)
- 2 scoops: \( \binom{32}{2} = 496 \)
- 3 scoops: \( \binom{33}{3} = 5456 \)
- 4 scoops: \( \binom{34}{4} = \frac{34 \cdot 33 \cdot 32 \cdot 31}{4 \cdot 3 \cdot 2 \cdot 1} = 46376 \)

✔ Total: \( 31 + 496 + 5456 + 46376 = 52359 \)

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Problem 5

> How many ways are there to arrange the four letters in the word MATH?

All letters are unique → permutation of 4 letters.

✔ Answer: \( 4! = 24 \)

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Problem 6

> If 8 people are running in a race, how many possible ways are there to place them in first, second, and third?

Order matters → permutation of 8 taken 3 at a time.

✔ Answer: \( P(8,3) = 8 \cdot 7 \cdot 6 = 336 \)

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Problem 7

> In a restaurant, how many ways are you select three different side dishes from eight available?

Order doesn’t matter → combination.

✔ Answer: \( \binom{8}{3} = 56 \)

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Problem 8

> Some pizzas have cheese placed on five numbers. If these slices do not want to use the same cheese on adjacent slices, how many ways can they do this if there are 3 cheeses available?

This is tricky. It says “five numbers” — likely means five slices arranged in a circle (since it mentions “adjacent slices”). We’re assigning one of 3 cheeses to each slice such that no two adjacent slices have the same cheese.

This is a classic circular arrangement coloring problem.

For a circle of n positions, with k colors, no two adjacent same color:

Formula: \( (k-1)^n + (-1)^n (k-1) \)

Here, n=5, k=3

✔ Answer: \( (3-1)^5 + (-1)^5 (3-1) = 32 + (-1)(2) = 32 - 2 = 30 \)

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Problem 9

> The city has a new license plate regulation. The last 12 digits must be odd. How many different groupings of 3 digits can be made from the available digits?

Wait — “last 12 digits must be odd”? That seems like a typo. Probably meant “the last digit must be odd” or “digits must be odd”.

But then it says: “How many different groupings of 3 digits can be made from the available digits?”

Assuming it means: how many 3-digit sequences can be made using only odd digits?

Odd digits: 1, 3, 5, 7, 9 → 5 choices.

If digits can repeat: \( 5^3 = 125 \)

If no repetition: \( P(5,3) = 5 \cdot 4 \cdot 3 = 60 \)

The problem says “groupings” — which might imply combinations, but since it’s about license plates, order matters.

Also, “available digits” — probably means digits 0–9, but “must be odd” — so only 1,3,5,7,9.

✔ Assuming repetition allowed: 125

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Problem 10

> A dog number of the state of Texas has a thousand's digit, a hundred's digit, and a ten's digit. How many different dog numbers can be made if no digit can be repeated?

This is confusingly worded. Probably means: a 3-digit number (thousands, hundreds, tens — wait, that’s 3 digits? Maybe it’s a 3-digit code).

Assume it’s a 3-digit number (hundreds, tens, units), and no digit repeats.

Digits 0–9, but first digit ≠ 0.

So:

- Hundreds digit: 1–9 → 9 choices
- Tens digit: 0–9 except hundreds digit → 9 choices
- Units digit: 8 choices left

✔ Answer: \( 9 \cdot 9 \cdot 8 = 648 \)

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Problem 11

> Part two of each of the Harry Potter books. Each book is clearly different from the others. How many different ways can you read the books if you must read them in order?

Wait — “Part two of each”? That doesn’t make sense. Probably meant:

> There are 7 Harry Potter books. Each is clearly different. How many different ways can you read them if you must read them in order?

If you must read them in order, then there’s only 1 way.

✔ Answer: 1

*(If it meant “in any order”, it would be 7! = 5040, but it says “must read them in order”)*

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Problem 12

> Part two of each of the Harry Potter books. Each book is clearly different from the others. How many different ways can you read the books if you can read them in any order?

Now it makes sense — assuming 7 books.

✔ Answer: \( 7! = 5040 \)

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Problem 13

> A group of three CDs has been nominated. In class of 25 students, how many different groups of students can be chosen to receive these CDs?

This is ambiguous.

Option 1: The CDs are identical → choose 3 students out of 25 → \( \binom{25}{3} = 2300 \)

Option 2: The CDs are different → assign 3 distinct CDs to 3 different students → \( P(25,3) = 25 \cdot 24 \cdot 23 = 13800 \)

The problem says “a group of three CDs” — suggests they are indistinct. Also, “groups of students” implies unordered selection.

✔ Answer: \( \binom{25}{3} = 2300 \)

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## ✔ Final Answers Summary:

1. 6
2. 2300
3. 5983
4. 52359
5. 24
6. 336
7. 56
8. 30
9. 125 *(assuming repetition allowed)*
10. 648
11. 1
12. 5040
13. 2300

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Let me know if you’d like any of these explained in more detail!