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. 3600 B. 3720 C. 3800 D. 3500
ANSWER
Verified
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.
Complete step-by-step answer: We have given the number of dyes with respect to each colour. Observe that for each dye, we have two cases regardless of its colour and whether they are chosen or not. So, the number of ways by which at least 1 green dye is chosen from 1 different green dye i.e Total number of selecting dyes -no selection of dyes 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 ${2}^{5}$-$1$ Similarly, for blue dyes 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. Now, these events are independent from each other. So, total number of ways will be \[ = ({2^5} - 1)({2^4} - 1){2^3} = 31 \times 15 \times 8 = 3720\] Hence, option B is correct.
Note: Events are independent from each other means, what dye we have got from a set of green dyes, it has nothing to do with what we are going to get from blue or red dyes. That is why we multiplied in order to get the final answer.