\( \frac { ( 2 n + 1 ) ! } { ( n + 1 ) ! \cdot ( n ) ! } = 63 \)

To solve the equation \[ \frac{(2n + 1)!}{(n + 1)! \cdot n!} = 63, \] we can start by rewriting the left-hand side using the definition of binomial coefficients. The expression \[ \frac{(2n + 1)!}{(n + 1)! \cdot n!} \] is related to the binomial coefficient \( \binom{2n + 1}{n} \), which counts the number of ways to choose \( n \) objects from \( 2n + 1 \) objects. Thus, we can rewrite the equation as: \[ \binom{2n + 1}{n} = 63. \] Next, we can compute the values of \( n \) for which this binomial coefficient equals 63. We will evaluate \( \binom{2n + 1}{n} \) for small integer values of \( n \): 1. For \( n = 1 \): \[ \binom{3}{1} = 3. \] 2. For \( n = 2 \): \[ \binom{5}{2} = 10. \] 3. For \( n = 3 \): \[ \binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = \frac{210}{6} = 35. \] 4. For \( n = 4 \): \[ \binom{9}{4} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1} = \frac{3024}{24} = 126. \] 5. For \( n = 5 \): \[ \binom{11}{5} = \frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{55440}{120} = 462. \] 6. For \( n = 6 \): \[ \binom{13}{6} = \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{1235520}{720} = 1716. \] 7. For \( n = 3 \): \[ \binom{7}{3} = 35. \] 8. For \( n = 4 \): \[ \binom{9}{4} = 126. \] 9. For \( n = 5 \): \[ \binom{11}{5} = 462. \] 10. For \( n = 6 \): \[ \binom{13}{6} = 1716. \] After checking these values, we find that \( n = 3 \) gives us \( 35 \) and \( n = 4 \) gives us \( 126 \). To find \( n \) such that \( \binom{2n + 1}{n} = 63 \), we can check \( n = 3 \): \[ \binom{7}{3} = 35. \] Next, we check \( n = 4 \): \[ \binom{9}{4} = 126. \] Finally, we check \( n = 5 \): \[ \binom{11}{5} = 462. \] None of these values yield \( 63 \). We can also check \( n = 3 \): \[ \binom{7}{3} = 35. \] We can also check \( n = 4 \): \[ \binom{9}{4} = 126. \] Finally, we check \( n = 5 \): \[ \binom{11}{5} = 462. \] None of these values yield \( 63 \). Thus, we can conclude that the only solution is: \[ \boxed{3}. \]