Math, asked by sgokuahulrai7080, 1 year ago

Given 5 different green dyes,4 different blue dyes and 3 different red dyes how many combination of dyes can be chosen, taking at least 1 green and at least 1 blue dye

Answers

Answered by Anonymous
8
Green dyes = 5

Blue dyes = 4

Red dyes = 3

No. of ways of choosing at least one green and at least one blue

 = (2^(5) - 1) (2^(4) - 1)(2^3)

= 3720
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