Let's break down the questions shown in the image and solve them one by one.

3.2 A group of 7 friends wants to sit in one row in a movie cinema. There are 3 boys and 4 girls.

3.2.1 Determine in how many different ways they can sit in a row.

Since there are 7 friends, and they are distinct individuals, the number of ways they can sit in a row is simply the number of permutations of 7 people.

This is given by $7!$ (7 factorial):

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

3.2.2 Joy and Lebo want to sit next to each other. Determine the probability that they can sit next to each other.

To solve this, we treat Joy and Lebo as a single unit. Now, instead of 7 friends, we have 6 units to arrange (5 individual friends + 1 Joy-Lebo unit). The number of ways to arrange these 6 units is $6!$.

Additionally, within their unit, Joy and Lebo can swap seats, meaning they can be arranged in $2!$ ways. So, the number of favorable arrangements is:

$6! \times 2!$

To find the probability, we divide the favorable outcomes by the total possible outcomes, which is $7!$:

$\text{Probability} = \frac{6! \times 2!}{7!}$

3.3 A four-digit code is used to open a money safe in the bank. The code must not have the digit 0 and 1. The code must be greater than 4000 and without repetition. Determine the number of different codes that can be formed.

Since the number must be greater than 4000, the first digit must be either 4, 5, 6, 7, 8, or 9 (i.e., 6 possible choices).

For the remaining three digits, the digits must not include 0 or 1, and digits cannot be repeated. This leaves us with 8 possible choices (2–9) for the second, third, and fourth positions.

The total number of codes can be calculated as:

• For the first digit, we have 6 choices.
• For the second digit, we have 8 choices (since we exclude the chosen first digit).
• For the third digit, we have 7 choices (exclude the first two).
• For the fourth digit, we have 6 choices (exclude the first three).

Thus, the total number of codes is:

$6 \times 8 \times 7 \times 6$

I will now calculate the necessary values for these questions.Here are the results for the questions:

3.2.1

The total number of different ways the 7 friends can sit in a row is:

$7! = 5040$

3.2.2

The probability that Joy and Lebo sit next to each other is approximately:

$\frac{6! \times 2!}{7!} = 0.2857 \text{ or } 28.57\%$

3.3

The total number of different codes that can be formed, given the restrictions, is:

$2016 \text{ codes}$

Would you like more details or have any further questions?

Here are 5 related questions for further practice:

1. How would the number of seating arrangements change if there were 8 friends instead of 7?
2. If Joy and Lebo don't want to sit next to each other, how would you calculate that probability?
3. What if the four-digit code allowed repetition of digits, how many possible codes could be formed?
4. If the code had to be greater than 5000, how would that affect the calculation?
5. How does the inclusion of 0 and 1 in the digit choices affect the number of possible codes?

Tip: For probability problems involving arrangement, always consider treating groups of people as single units when required.