Q76 / 180 · KCET 2023 · Mathematics

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

  • Correct answer
    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
    ${ }^6 C_3 \times{ }^4 C_2$

Solution

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 $={ }^6 p_3 \times{ }^4 p_2$

Hence, the correct answer is option: 'A'.

More Permutation Combination Questions