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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 at least one green and one blue dye is

$\begin{array}{1 1}(A)\;3600\\(B)\;3720\\(C)\;3800\\(D)\;3600\end{array} $

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A)
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Green dyes =5
Blue dyes =4
Red dyes =3
The no of ways of choosing of at least one green and one blue is =$(2^5-1)\times (2^4-1)\times 2^3$
$\Rightarrow (32-1)\times (16-1)\times 8$
$\Rightarrow 31\times 15\times 8$
$\Rightarrow 3720$
Hence (B) is the correct answer.
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