(Q). Given 6 different green dyes, 5 different blue dyes & 4 different red dyes, the number of combination of dyes which can be chosen taking at least one green & one blue dye is?
Solution:-
one or more (at least one) green dye can be chosen in ⁶C₁ + ⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆ = (2⁶ – 1) ways
One or more (at least one) blue dye can be chosen in ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅ = (2⁵ – 1) ways
Zero or more (Any number of) red dye can be chosen in ⁴C₀ + ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄ = 2⁴ ways
So, total number of required selection
= (2⁶ – 1)×(2⁵ – 1)×2⁴
= 63×31×16
= 31248 ways Answer