KCET · Maths · Permutation Combination
Ten chairs are numbered as 1 to 10 . Three women and two men wish to occupy one chair each. First the women choose the chairs marked 1 to 6 , then the men choose the chairs from the remaining. The number of possible ways is
- A \({ }^6 P_3 \times{ }^4 P_2\)
- B \({ }^6 C_3 \times{ }^4 P_2\)
- C \({ }^6 P_3 \times{ }^4 C_2\)
- D \({ }^6 C_3 \times{ }^4 C_2\)
Answer & Solution
Correct Answer
(A) \({ }^6 P_3 \times{ }^4 P_2\)
Step-by-step Solution
Detailed explanation
We have, 3 women and 2 men and first
women choose the chairs marked 1 to 6 .
\(\because\) Total number of chair \(=6\)
So, the number of arrangement \(={ }^6 P_3\)
Now, 2 men choose from the remaining 4 chairs.
So, the number of arrangement \(={ }^4 p_2\) ways
\(\therefore\) Total number of arrangements \(={ }^6 p_3 \times{ }^4 p_2\)
women choose the chairs marked 1 to 6 .
\(\because\) Total number of chair \(=6\)
So, the number of arrangement \(={ }^6 P_3\)
Now, 2 men choose from the remaining 4 chairs.
So, the number of arrangement \(={ }^4 p_2\) ways
\(\therefore\) Total number of arrangements \(={ }^6 p_3 \times{ }^4 p_2\)
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