Three men have `4` coats, `6` waistcoats and `5` caps. In how many ways can they wear them?

Options

A. `""^(4)P_(3).""^(6)P_(3).""^(5)P_(3)(3!)^3`
B. `""^(4)C_(3).""^(6)C_(3).""^(5)C_(3)`
C. `""^(4)C_(3).""^(6)C_(3).""^(5)C_(3).6!.4!.5!`
D. `""^(4)P_(3).""^(6)P_(3).""^(5)P_(3)`

Answer

D

Explanation

Total number of ways in which 3 men can wear 4 coats
= number of arrangements of 4 coats taking 3 at a time
= ` ""^(4)P_(3)=(4xx3xx2) =24.`
Total number of ways in which 3 men can wear 6 waistcoats
= number of arrangements of 6 waistcoats taking 3 at a time
= ` ""^(6)P_(3) =(6xx5xx4) =120.`
Total number of ways in which 3 men can wear 5 caps
= ` ""^(5)P_(3) =(5xx4xx3)=60.`
Hence, by the fundamental principal of multiplication, the required number of ways = `(24xx120xx60)` = 172800.