Question

A student is to answer \(10\) out of \(13\) questions in an examination such that he must choose at least \(4\) from the first \(5\) questions. The number of choices available to him is:

• A. \(196\)
• B. \(140\)
• C. \(168\)
• D. \(176\)

Hint:

For selection problems with “at least” conditions, split the problem into separate cases and add the number of ways from all valid cases.

Answer 1

The Correct Option is A

Step 1: Understand the selection condition.
There are total \(13\) questions.
The student has to answer \(10\) questions.
Among the first \(5\) questions, he must choose at least \(4\).
So, there are two possible cases:
\[ \text{Case 1: Choose }4\text{ questions from first }5 \] \[ \text{Case 2: Choose }5\text{ questions from first }5 \]

Step 2: Case 1: Choose \(4\) from first \(5\).
Number of ways to choose \(4\) questions from first \(5\) is \[ {}^5C_4. \] Now, total questions to be selected are \(10\).
Already selected questions are \(4\).
So, remaining questions to be selected are \[ 10-4=6. \] These \(6\) questions must be chosen from the remaining \(8\) questions.
Thus, number of ways in this case is \[ {}^5C_4 \times {}^8C_6. \] \[ =5\times 28 \] \[ =140. \]

Step 3: Case 2: Choose \(5\) from first \(5\).
Number of ways to choose all \(5\) questions from first \(5\) is \[ {}^5C_5. \] Now, remaining questions to be selected are \[ 10-5=5. \] These \(5\) questions must be chosen from the remaining \(8\) questions.
Thus, number of ways in this case is \[ {}^5C_5 \times {}^8C_5. \] \[ =1\times 56 \] \[ =56. \]

Step 4: Add both cases.
Total number of choices is \[ {}^5C_4{}^8C_6+{}^5C_5{}^8C_5. \] \[ =140+56 \] \[ =196. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{196} \]