QUESTION

Given 5 different green dyes, 4 different blue dyes and 3 different red dyes. The number of combinations of dyes which can be chosen taking atleast one green and one blue dye is?A. 3600B. 3720C. 3800D. 3500

Hint: For each dye we have two cases. It’ll be chosen or not.So the number of ways of selecting the dyes will gives combination.Using this concept to take atleast one green dye from 5 different dyes,First we have to calculate number of ways of selecting 5 different green dyes it is given by ${2}^{5}$ then we have to subtract one way in which there is no selection of dye i.e ${2}^{5}$-$1$.Similarly calculate for blue dye and number of ways of selecting red dye will be ${2}^{3}$ ways, as these events are independent to each other the product of all these value gives required answer.

Total number of combinations of selecting different green dyes is ${2}^{5}$
Out of combinations there is one way in which there is no selection of dye So ,it is given as ${2}^{5}$-$1$
Total number of combinations of selecting different blue dyes is ${2}^{4}$
Out of combinations there is one way in which there is no selection of dye So ,it is given as ${2}^{4}$-$1$
and Three different red dye can be chosen in ${2^3}$ ways.
So, total number of ways will be $= ({2^5} - 1)({2^4} - 1){2^3} = 31 \times 15 \times 8 = 3720$