The problem in the image is as follows:

Ratna is taking a test consisting of 10 questions. She is allowed to answer 7 questions, but with the condition that questions 1 and 2 must be answered. The question asks how many different ways she can choose the remaining questions under this condition.

Solution:

Since Ratna must answer questions 1 and 2, this leaves her needing to choose 5 more questions out of the remaining 8 (since 2 questions are already fixed).

• Total questions = 10
• Questions already chosen (1 and 2) = 2
• Remaining questions = $10 - 2 = 8$
• Questions to choose from the remaining = 5

Thus, the number of ways to choose 5 questions from the 8 remaining ones is given by the combination formula:

$\binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!}$

Let me calculate that for you.The number of ways Ratna can choose 5 questions from the remaining 8 is 56.

Thus, there are 56 different ways for Ratna to select the remaining questions under the given conditions.

Would you like more details on any part of the explanation?

Here are some related questions for further practice:

1. What if Ratna needed to choose 6 questions, but questions 3 and 4 must be answered?
2. How many ways can Ratna choose 7 questions if there are no restrictions?
3. What if only one of the questions 1 or 2 must be answered, not both?
4. What if questions 1, 2, and 3 must all be answered, and Ratna can choose 4 more from the remaining 7?
5. How would the solution change if Ratna could only answer 6 questions?

Tip: Remember that combinations (denoted as $\binom{n}{r}$) are used when the order of selection does not matter, and they are very useful in counting problems.