text: ## Precalculus (6th Edition) Blitzer
  
  The required solution is $210$
  We know that the representation $_{n}{{P}_{r}}$ implies that the number of possible well-organized arrangements of n items is taken r at a time. And the number of possible well-organized arrangements of n items taken r at a time can be evaluated as: $_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ The provided expression is $_{7}{{P}_{3}}$. Here, $n=7,r=3$. Put the value of n, r in the above formula. Then: \begin{align} & _{7}{{P}_{3}}=\frac{7!}{\left( 7-3 \right)!} \\ & =\frac{7!}{4!} \\ & =\frac{7\cdot 6\cdot 5\cdot 4!}{4!} \end{align} And simplify further, \begin{align} & \frac{7\cdot 6\cdot 5\cdot 4!}{4!}=7\cdot 6\cdot 5 \\ & =210 \end{align} Hence, $_{7}{{P}_{3}}=210$
explanation: The document provides a solution to a permutation problem from a Precalculus textbook. It explains the formula for permutations and applies it to calculate \(_{7}P_{3}\), which results in 210. The document contains the answer to a specific math question but does not explicitly state the question itself.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ## Precalculus (6th Edition) Blitzer
  
  The required solution is $210$
  We know that the representation $_{n}{{P}_{r}}$ implies that the number of possible well-organized arrangements of n items is taken r at a time. And the number of possible well-organized arrangements of n items taken r at a time can be evaluated as: $_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ The provided expression is $_{7}{{P}_{3}}$. Here, $n=7,r=3$. Put the value of n, r in the above formula. Then: \begin{align} & _{7}{{P}_{3}}=\frac{7!}{\left( 7-3 \right)!} \\ & =\frac{7!}{4!} \\ & =\frac{7\cdot 6\cdot 5\cdot 4!}{4!} \end{align} And simplify further, \begin{align} & \frac{7\cdot 6\cdot 5\cdot 4!}{4!}=7\cdot 6\cdot 5 \\ & =210 \end{align} Hence, $_{7}{{P}_{3}}=210$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document provides a solution to a permutation problem from a Precalculus textbook. It explains the formula for permutations and applies it to calculate \(_{7}P_{3}\), which results in 210. The document contains the answer to a specific math question but does not explicitly state the question itself.
  
  Label: question-and-answer