Question

There are five different green dyes, four different blue dyes and three different red dyes. The total number of combinations of dyes that can be chosen taking at least one green and one blue dye is

• A. $3255$
• B. $2^{12}$
• C. $3720$
• D. $3755$

Answer 1

Answer: (C)

$3720$

No. of ways of choosing at least one green dye is

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

No. of ways of choosing at least one blue dye is

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

No. of ways of choosing any number of red dyes is

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

So total no. of ways = $31\times 15\times 8=3720$