Given 5 different green dyes, 4 different blue dyes and 3 different red dyes. The number of combinations of dyes that can be choosen by taking at least one green and one blue dye is

Number of ways of selecting at least one green dye out of 5 different green dyes.

${ }^5{C}_1{+}^5{C}_2{+}^5{C}_3{+}^5{C}_4{+}^5{C}_5=2^5-1$

Also at least one blue dye can be selected out of 4 blue dyes in

${ }^4{C}_1{+}^4{C}_2{+}^4{C}_3{+}^4{C}_4=2^4-1$

Again , 3 different red dyes can be selected in

${ }^3{C}_0{+}^3{C}_1{+}^3{C}_2{+}^3{C}_3=2^3$ ways

$\therefore$ Required ways $=(2^5-1)(2^4-1)2^3=3720$