Question

Given 5 different green dyes, four different blue dyes and three different red dyes, how many combinations of dyes can be chosen taking at least one green and one blue dye?

Answer 1

At. least one green dye can be selected out of 5 green dyes in $2^5$ - 1, i.e. in 31 ways. Similarly at least one blue dye can be selected out of 4 in $2^4$ - 1 in 15 ways. For red dyes there is no restriction; you may include it or not include it. Thus there are two ways of disposing of each of red dye. Thus the total number of selection of red dye is $2^3$ = 8
. Hence the required number of ways 31 x 15 x 8 = 3720.