Question 94329
*[Tex \LARGE \textrm{\frac{_{4}P_{3}}{_{8}P_{3}}] Start with the given expression




First lets evaluate the numerator: *[Tex \LARGE \textrm{_{4}P_{3}]




*[Tex \LARGE \textrm{_{n}P_{r}=]{{{n!/(n-r)!}}} Start with the given formula




*[Tex \LARGE \textrm{_{4}P_{3}=]{{{4!/(4-3)!}}} Plug in {{{n=4}}} and {{{r=3}}}




*[Tex \LARGE \textrm{_{4}P_{3}=]{{{4!/1!}}} Subtract {{{4-3}}} to get 1




Expand 4!
*[Tex \LARGE \textrm{_{4}P_{3}=]{{{(4*3*2*1)/1!}}}




Expand 1!
*[Tex \LARGE \textrm{_{4}P_{3}=]{{{(4*3*2*1)/(1)}}}




*[Tex \LARGE \textrm{_{4}P_{3}=]{{{(4*3*2*cross(1))/(cross(1))}}}  Cancel




*[Tex \LARGE \textrm{_{4}P_{3}=]{{{4*3*2}}}  Simplify





*[Tex \LARGE \textrm{_{4}P_{3}=]{{{24}}}  Now multiply 4*3*2 to get 24






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Now lets evaluate the denominator: *[Tex \LARGE \textrm{_{8}P_{3}]





*[Tex \LARGE \textrm{_{n}P_{r}=]{{{n!/(n-r)!}}} Start with the given formula




*[Tex \LARGE \textrm{_{8}P_{3}=]{{{8!/(8-3)!}}} Plug in {{{n=8}}} and {{{r=3}}}




*[Tex \LARGE \textrm{_{8}P_{3}=]{{{8!/5!}}} Subtract {{{8-3}}} to get 5




Expand 8!
*[Tex \LARGE \textrm{_{8}P_{3}=]{{{(8*7*6*5*4*3*2*1)/5!}}}




Expand 5!
*[Tex \LARGE \textrm{_{8}P_{3}=]{{{(8*7*6*5*4*3*2*1)/(5*4*3*2*1)}}}




*[Tex \LARGE \textrm{_{8}P_{3}=]{{{(8*7*6*cross(5*4*3*2*1))/(cross(5*4*3*2*1))}}}  Cancel




*[Tex \LARGE \textrm{_{8}P_{3}=]{{{8*7*6}}}  Simplify





*[Tex \LARGE \textrm{_{8}P_{3}=]{{{336}}}  Now multiply 8*7*6 to get 336



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So *[Tex \LARGE \textrm{\frac{_{4}P_{3}}{_{8}P_{3}}=\frac{24}{336}]



Now reduce {{{24/336}}} to get {{{1/14}}}





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Answer:



So    *[Tex \LARGE \textrm{\frac{_{4}P_{3}}{_{8}P_{3}}=\frac{1}{14}]