## 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, $_{7}{{P}_{3}}=210$
The text passage provides a solution to a permutation problem, specifically calculating $_{7}{{P}_{3}}$. The question is implicitly stated through the provided expression, and the answer is given as 210.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
## 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, $_{7}{{P}_{3}}=210$
user
The text passage provides a solution to a permutation problem, specifically calculating $_{7}{{P}_{3}}$. The question is implicitly stated through the provided expression, and the answer is given as 210.

TYPE: SINGLE_QUESTION_WITH_ANSWER