\( 1 \leftarrow \quad \begin{array}{l}\text { Decide whether each of the following statements is true. If false, demonstrate why. } \\ \text { a. } 9!=9 \cdot 8! \\ \text { b. } 6!+3!=9! \\ \text { c. } \frac{9!}{3!}=3! \\ \text { a. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. } \\ \text { A. The statement is true. } \\ \text { B. The statement is false because } 9!=\square \text { and } 9 \cdot 8!=\square \\ \text { (Simplify your answers.) }\end{array} \)

The first statement, \( 9! = 9 \cdot 8! \), is indeed true. This is because the factorial of 9 (denoted as \( 9! \)), which is \( 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \), can be rewritten as \( 9 \times (8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) \), which is precisely \( 9 \cdot 8! \). As for the second statement, \( 6! + 3! = 9! \), this one is false. Calculating the left side, we find \( 6! = 720 \) and \( 3! = 6 \), so \( 720 + 6 = 726 \). On the right side, \( 9! \) equals \( 362880 \). Clearly, \( 726 \neq 362880 \). Lastly, the third statement, \( \frac{9!}{3!} = 3! \), is also false. When we calculate \( \frac{9!}{3!} \), we essentially get \( \frac{362880}{6} = 60480 \), but \( 3! = 6 \). Thus, \( 60480 \neq 6 \).