KCET · Maths · Permutation Combination
A student has to answer 10 questions, choosing at least 4 from each of the parts \(A\) and \(B\). If there are 6 questions in part \(A\) and 7 in part \(B\), then the number of ways can the student choose 10 questions is
- A 256
- B 352
- C 266
- D 426
Answer & Solution
Correct Answer
(C) 266
Step-by-step Solution
Detailed explanation
Given,
Total number of questions in part \(A=6\)
Total number of questions in part \(B=7\)
Pattern to choose 10 questions from 13 questions is
4 from part \(A\) and 6 from part \(B\) or
5 from part \(A\) and 5 from part \(B\) or
6 from part \(A\) and 4 from part \(B\).
\(\therefore\) Total required number of ways
\(\begin{aligned}
&=\left({ }^{6} C_{4} \times{ }^{7} C_{6}\right)+\left({ }^{6} C_{5} \times{ }^{7} C_{5}\right)+\left({ }^{6} C_{6} \times{ }^{7} C_{4}\right) \\
&=(15 \times 7)+(6 \times 21)+(1 \times 35) \\
&=105+126+35=266
\end{aligned}\)
Total number of questions in part \(A=6\)
Total number of questions in part \(B=7\)
Pattern to choose 10 questions from 13 questions is
4 from part \(A\) and 6 from part \(B\) or
5 from part \(A\) and 5 from part \(B\) or
6 from part \(A\) and 4 from part \(B\).
\(\therefore\) Total required number of ways
\(\begin{aligned}
&=\left({ }^{6} C_{4} \times{ }^{7} C_{6}\right)+\left({ }^{6} C_{5} \times{ }^{7} C_{5}\right)+\left({ }^{6} C_{6} \times{ }^{7} C_{4}\right) \\
&=(15 \times 7)+(6 \times 21)+(1 \times 35) \\
&=105+126+35=266
\end{aligned}\)
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