Combinatorics: More than One Scenario - Calculate Separately and Add (OR relations)
A photographer is to take group photographs of a class of students for the school magazine such that each photograph should have five students. If there are four girls and four boys in the class and each photograph must not have two girls or two boys standing next to each other, how many different photographs can the photographer take?
Correct
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Let's start with the photographs with girls and boys standing in the order G-B-G-B-G. Since there's no repetition, the number of options for each successive child from the same gender decreases. For example, if there are 4 choices for the first girl, then there are only 3 choices for the next girl (third place), and 2 choices for last girl (fifth place). The same goes for boys (second and fourth places).
>$$\mathop{\fbox{${4}$}}\limits^{{}{}\hspace{0.33em}{\text{G}}}
\hspace{0.33em}\times\mathop{\fbox{${4}$}}\limits^{{}{}\hspace{0.33em}
{\text{B}}}\hspace{0.33em}\times \mathop{\fbox{${3}$}}\limits^{{}{}\hspace{0.33em}{\text{G}}}
\hspace{0.33em}\times\mathop{\fbox{${3}$}}\limits^{{}{}\hspace{0.33em}{\text{B}}}\hspace{0.33em}
\times\mathop{\fbox{${2}$}}\limits^{{}{}\hspace{0.33em}{\text{G}}}={288}$$
Now, do the same for photographs with girls and boys standing in the order B-G-B-G-B. Note that intuitively, this scenario should have the same number of permutations, as we have an identical 4 boys and 4 girls to choose from.
>$$\mathop{\fbox{${4}$}}\limits^{{}{}\hspace{0.33em}{\text{B}}}
\hspace{0.33em}\times\mathop{\fbox{${4}$}}\limits^{{}{}\hspace{0.33em}
{\text{G}}}\hspace{0.33em}\times \mathop{\fbox{${3}$}}\limits^{{}{}\hspace{0.33em}{\text{B}}}
\hspace{0.33em}\times\mathop{\fbox{${3}$}}\limits^{{}{}\hspace{0.33em}{\text{G}}}\hspace{0.33em}
\times\mathop{\fbox{${2}$}}\limits^{{}{}\hspace{0.33em}{\text{B}}}={288}$$
Since there is an "or" relationship here between either of the two types of photographs that can be taken, add the combinations.
>$$\text{Arrangements} = 288 + 288 = 576$$
Incorrect.
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Incorrect.
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Incorrect.
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Incorrect.
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$$80$$
$$288$$
$$4^4$$
$$576$$
$$288^2$$
