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A box contains 4 different black balls 3 different red balls and 5 different blue balls. In how many ways can the balls be selected if every selection must have at least 1 black ball and one red ball?

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$(2^{4} - 1) (2^{3} - 1) 2^{5}$

$2^{12}$

$2^{12} - 1$

$(2^{4} - 1) (2^{3} - 1) (2^{5} - 1)$

answer is B.

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Detailed Solution

4 black

$^{4} C_{1} +^{4} C_{2} +^{4} C_{3} +^{4} C_{4}$

$= 2^{4} - 1$

3 red

$^{3} C_{1} +^{3} C_{2} +^{3} C_{3}$

$= 2^{3} - 1$

5 blue

$^{5} C_{0} +^{5} C_{1} +^{5} C_{2} +^{5} C_{3} +^{5} C_{4} +^{5} C_{5}$

$2^{5}$

$(2^{4} - 1) (2^{3} - 1) 2^{5}$

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