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 atleast one green and one blue dye is

• A. 3600
• B. 3720
• C. 3800
• D. 3500

Answer 1

Answer: (B)

3720

Answer 2

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 $31 \times 15 \times 8=3720$.