Given 5 different green dyes, four different blue dyes and three different red dyes,

Question:

Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is

(a) 3600

(b) 3720

(c) 3800

(d) 3600

Solution:

Number of ways of selecting one green dye is 5C1 + 5C2 + 5C5C4 + 5C5  = 25−1 ways. 

Number of ways of selecting one blue dye can be choosen is 4C1 + 4C2 + 4C4C4 = 24−1 ways.

and 

Number of ways of selecting red dye can be choosen in 3C0 + 3C1 + 3C3C3 = 23 ways.

So, total number of required selection 

= (25 − 1) × (24−1) × 23

= 3720

Hence, the correct answer is option B.

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