[tex] \underline{\underline{\large{\red{\mathcal{ ✒ GIVEN:}}}}} [/tex]
[tex]\bm{A.}[/tex]
[tex]\bullet \: \: \rm{4 \: boys}[/tex]
[tex]\bullet \: \: \rm{4 \: girls}[/tex]
[tex]\bm{B.}[/tex]
[tex]\bullet \: \: \rm{2 \: paticular \: boys}[/tex]
[tex] \underline{\underline{\large{\red{\mathcal{REQUIRED:}}}}} [/tex]
How many different ways can they be seated if A?
[tex] \underline{\underline{\large{\red{\mathcal{SOLUTION:}}}}} [/tex]
In this question, respectively, 4 boys and 4 girls are to be seated around a circular table. The following questions include whether they can sit anywhere, how they can possibly arrange themselves, and whether, as two boys, they wish to sit together. Among circular permutations, the number of ways in which n objects can be arranged in a circle is (n-1)! ways. Because the problem deals with 8 people, out of whom four are boys and four are girls:
[tex]\tt{(n-1)! = (8-1)! = 7!}[/tex]
[tex]\tt{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}[/tex]
[tex]\large{\tt{\purple{=5,040}}}[/tex]
When a decision has to be made regarding two particular boys as the unit of the block, although the numbers are reduced by one person because of the relationship these two boys have with each other, therefore, we will have 7 people to put around the table (properly, the group of 2 boys included the other 6). Besides, there are two ways to place both of the two boys in relation to the block identified above. So the number of arrangements is given by:
[tex]\boxed{\large{\bm{\red{2 \cdot (n'-1)!=}}}}[/tex]
[tex]\tt{2 \cdot (7-1)!}[/tex]
[tex]\tt{2 \cdot 6! =}[/tex]
[tex]\tt{6 \times 5 \times 4 \times 3 \times 2 \times 1=720}[/tex]
[tex]\tt{2 \cdot 720 }[/tex]
[tex]\large{\tt{\purple{=1,140}}}[/tex]
Final Answer:
Therefore the answers are:
[tex]\bm{A.}[/tex]
[tex]\large{\tt{\purple{5,040}}}[/tex]
[tex]\bm{B.}[/tex]
[tex]\large{\tt{\purple{=1,140}}}[/tex]
