Question

Given $5$ different green dyes, $4$ and different blue dyes and $3$ different red dyes. The number of combinations of dyes can be chosen taking at least one green and one blue dyes is:

• A. $3870$
• B. $3715$
• C. $3720$
• D. $3876$

Answer 1

The correct option is C $3720$

The number of ways of choosing of at least one green and one blue is $=(2^5-1)\times (2^4-1)\times 2^3$

$=(32-1)\times (16-1)\times 8$

$=31\times 15\times 8$

$=3720$