Question

A student has to answer 10 out of 13 questions in an examination choosing atleast 3 from the 5 particular questions. The number of choices available to the student is

• A. \( 196 \)
• B. \( 276 \)
• C. \( 326 \)
• D. \( 156 \)

Hint:

When computing combinations like \( \binom{8}{6} \), always use the symmetry property \( \binom{n}{r} = \binom{n}{n-r} \). Thus, \( \binom{8}{6} = \binom{8}{2} = \frac{8 \times 7}{2} = 28 \), which saves valuable time!

Answer 1

The Correct Option is B

Step 1: Split the questions into groups.
There are 13 questions in total. We treat 5 of them as the special group and the other $13-5=8$ as the rest.
Step 2: Read the rule.
The student answers 10 questions and must pick at least 3 from the special 5. Since only 5 are special, the picks from there can be 3, 4, or 5.
Step 3: Case of 3 special.
Choose 3 from 5 and 7 from 8: $\binom{5}{3}\binom{8}{7} = 10 \times 8 = 80$.
Step 4: Case of 4 special.
Choose 4 from 5 and 6 from 8: $\binom{5}{4}\binom{8}{6} = 5 \times 28 = 140$.
Step 5: Case of 5 special.
Choose 5 from 5 and 5 from 8: $\binom{5}{5}\binom{8}{5} = 1 \times 56 = 56$.
Step 6: Add the cases.
Total $= 80 + 140 + 56$. \[ \boxed{276} \]