Let's break down the questions from the image and solve them one by one:

Question 3a

Problem: In a math test, students are required to answer 4 out of 7 questions. How many ways can the questions be chosen if there are no restrictions?

Solution: This is a combinations problem where the order of selection doesn't matter. The formula for combinations is:

$C(n, k) = \frac{n!}{k!(n-k)!}$

Where:

• $n$ is the total number of questions (7),
• $k$ is the number of questions to choose (4).

So the number of ways to choose 4 questions from 7 is:

$C(7, 4) = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!}$

Let me calculate this.The number of ways to choose 4 questions out of 7 without any restrictions is 35.

Question 3b

Problem: Find the number of ways in which the questions can be chosen if the last question is compulsory.

Solution: If the last question is compulsory, then we are left to choose 3 more questions from the remaining 6 questions. This again is a combinations problem, but now we are choosing 3 questions from 6.

The number of ways to do this is:

$C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}$

Let me calculate this next.The number of ways to choose 3 additional questions (when the last question is compulsory) from the remaining 6 is 20.

Would you like further explanation or have any other questions?

Here are 5 related questions to consider:

1. How would the solution change if we had to choose 5 questions out of 7 instead?
2. How do the formulas for combinations and permutations differ in application?
3. What happens if there were restrictions on choosing certain questions?
4. Can the combination formula be applied to scenarios outside of math tests? How?
5. What is the difference between solving combinations with and without compulsory selections?

Tip: Remember that combinations are useful when order doesn't matter. Always check if the problem is asking for a specific order (which would require permutations instead).