Ten chairs are numbered from 1 to 10. Three women and two men wish to occupy one chair each. First, the women choose chairs numbered from 1 to 6, then the men choose chairs from the remaining chairs. The number of possible ways is

Question

$\binom{6}{3} \times \mathrm{P}(4,2)$

$\mathrm{P}(6,3) \times \binom{4}{2}$

$\binom{6}{3} \times \binom{4}{2}$

$\mathrm{P}(6,3) \times \mathrm{P}(4,2)$

Solution

Please login to view the detailed solution steps...

Go to DASH