Combinatorics quiz worksheet with multiple-choice questions on permutations, probability, and arrangements.

Let's solve each problem step by step.

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Problem 1:

Gloria and Bob and five other people walk through a doorway one at a time. In how many ways can the seven people go through the doorway if Bob goes through the doorway after Gloria?

#### Solution:
1. Total number of people: There are 7 people in total.
2. Total arrangements without restriction: The total number of ways to arrange 7 people is given by \(7!\):
\[
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
\]
3. Condition: Bob must go through the doorway after Gloria. For any pair of people (Gloria and Bob), there are exactly two possible orders: Gloria before Bob or Bob before Gloria. Since these two orders are equally likely, exactly half of the total arrangements will have Gloria before Bob.
4. Number of valid arrangements: Half of the total arrangements:
\[
\frac{7!}{2} = \frac{5040}{2} = 2520
\]

#### Final Answer:
\[
\boxed{B}
\]

---

Problem 2:

A child arranges 5 blocks in a row. Two blocks are painted blue, one is painted red, one is painted green, and one is painted white. How many possible patterns of colours are there if the two blue blocks are next to one another?

#### Solution:
1. Treat the two blue blocks as a single "super block": Since the two blue blocks must be next to each other, we can treat them as a single unit. This reduces the problem to arranging 4 units: the "super block" (BB), the red block (R), the green block (G), and the white block (W).
2. Arrange the 4 units: The number of ways to arrange these 4 units is given by \(4!\):
\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]
3. Internal arrangement of the blue blocks: Within the "super block," the two blue blocks can be arranged in only 1 way (since they are identical).

#### Final Answer:
\[
\boxed{B}
\]

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Problem 3:

Mr. and Mrs. Brown and their four children go to the cinema. They are randomly allocated six adjacent seats in a single row. What is the probability that the four children are allocated seats next to each other?

#### Solution:
1. Total number of arrangements: The total number of ways to arrange 6 people in a row is given by \(6!\):
\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]
2. Treat the four children as a single "super block": If the four children must sit next to each other, we can treat them as a single unit. This reduces the problem to arranging 3 units: the "super block" (the four children), Mr. Brown, and Mrs. Brown.
3. Arrange the 3 units: The number of ways to arrange these 3 units is given by \(3!\):
\[
3! = 3 \times 2 \times 1 = 6
\]
4. Arrange the four children within the "super block": The four children can be arranged among themselves in \(4!\) ways:
\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]
5. Total favorable arrangements: The total number of favorable arrangements is the product of the number of ways to arrange the 3 units and the number of ways to arrange the children within the "super block":
\[
3! \times 4! = 6 \times 24 = 144
\]
6. Probability: The probability that the four children are seated next to each other is the ratio of the number of favorable arrangements to the total number of arrangements:
\[
\text{Probability} = \frac{\text{Favorable arrangements}}{\text{Total arrangements}} = \frac{144}{720} = \frac{1}{5}
\]

#### Final Answer:
\[
\boxed{A}
\]

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Problem 4:

How many arrangements can be made using all the letters of the word ISOSCELES?

#### Solution:
1. Identify the letters and their frequencies: The word "ISOSCELES" has 9 letters:
- I: 1
- S: 3
- O: 1
- C: 1
- E: 2
- L: 1
2. Formula for permutations of multiset: The number of distinct permutations of a multiset is given by:
\[
\frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}
\]
where \(n\) is the total number of items, and \(n_1, n_2, \ldots, n_k\) are the frequencies of the distinct items.
3. Apply the formula: Here, \(n = 9\), and the frequencies are \(1, 3, 1, 1, 2, 1\). Thus:
\[
\text{Number of arrangements} = \frac{9!}{1! \cdot 3! \cdot 1! \cdot 1! \cdot 2! \cdot 1!}
\]
4. Calculate the factorials:
\[
9! = 362880, \quad 3! = 6, \quad 2! = 2
\]
5. Simplify the expression:
\[
\frac{9!}{1! \cdot 3! \cdot 1! \cdot 1! \cdot 2! \cdot 1!} = \frac{362880}{1 \cdot 6 \cdot 1 \cdot 1 \cdot 2 \cdot 1} = \frac{362880}{12} = 30240
\]

#### Final Answer:
\[
\boxed{C}
\]

---

Problem 5:

Seven people are seated at a round table. What is the probability that two people, Jenny and Lindsay, do not sit next to one another?

#### Solution:
1. Total number of circular arrangements: For \(n\) people sitting around a round table, the number of distinct arrangements is \((n-1)!\). Here, \(n = 7\):
\[
(7-1)! = 6! = 720
\]
2. Arrangements where Jenny and Lindsay sit next to each other: Treat Jenny and Lindsay as a single "super block." This reduces the problem to arranging 6 units (the "super block" and the other 5 people). The number of ways to arrange these 6 units in a circle is \((6-1)! = 5!\):
\[
5! = 120
\]
Within the "super block," Jenny and Lindsay can switch places, so there are \(2!\) ways to arrange them:
\[
2! = 2
\]
Therefore, the total number of arrangements where Jenny and Lindsay sit next to each other is:
\[
5! \times 2! = 120 \times 2 = 240
\]
3. Arrangements where Jenny and Lindsay do not sit next to each other: Subtract the number of arrangements where they sit next to each other from the total number of arrangements:
\[
720 - 240 = 480
\]
4. Probability: The probability that Jenny and Lindsay do not sit next to each other is the ratio of the number of favorable arrangements to the total number of arrangements:
\[
\text{Probability} = \frac{\text{Favorable arrangements}}{\text{Total arrangements}} = \frac{480}{720} = \frac{2}{3}
\]

#### Final Answer:
\[
\boxed{A}
\]

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Final Answers:

1. \(\boxed{B}\)
2. \(\boxed{B}\)
3. \(\boxed{A}\)
4. \(\boxed{C}\)
5. \(\boxed{A}\)