Question

Given $5$ different green dyes, $4$ different blues dyes and $3$ different red dyes, the number of combinations of dyes that can be chosen by taking at least one green and one blue dye is

• A. $248$
• B. $120$
• C. $3720$
• D. $465$

Answer 1

Answer: (C)

$3720$

Number of ways of selecting atleast 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.

Thus required ways$=(2^5-1)(2^4-1)\left(2^3\right)=3720$