The largest value of `n in N` for which `(74)/(""^(n)P_(n)) gt (""^(n + 3)P_(3))/(""^(n +1)P_(n + 1))` is _______

Answer

To solve the inequality

\[ \frac{74}{nP_n} > \frac{(n + 3)P_3}{(n + 1)P_{n + 1}}, \]

we start by recalling the formula for permutations, which states that

\[ nP_r = \frac{n!}{(n - r)!}. \]

### Step 1: Rewrite the permutations

Using the permutation formula, we can rewrite both sides of the inequality:

- The left side becomes:
\[ nP_n = \frac{n!}{(n - n)!} = n!. \]
Thus, the left side is:
\[ \frac{74}{n!}. \]

- The right side becomes:
\[ (n + 3)P_3 = \frac{(n + 3)!}{(n + 3 - 3)!} = \frac{(n + 3)!}{n!} = (n + 3)(n + 2)(n + 1), \]
and
\[ (n + 1)P_{n + 1} = \frac{(n + 1)!}{(n + 1 - (n + 1))!} = (n + 1)!. \]
Thus, the right side simplifies to:
\[ \frac{(n + 3)(n + 2)(n + 1)}{(n + 1)!} = \frac{(n + 3)(n + 2)(n + 1)}{(n + 1) n!} = \frac{(n + 3)(n + 2)}{n!}. \]

### Step 2: Set up the inequality

Now we have the inequality:

\[ \frac{74}{n!} > \frac{(n + 3)(n + 2)}{n!}. \]

### Step 3: Eliminate the common term

Since \(n!\) is positive for \(n \in \mathbb{N}\), we can multiply both sides by \(n!\):

\[ 74 > (n + 3)(n + 2). \]

### Step 4: Expand and rearrange

Expanding the right side gives:

\[ 74 > n^2 + 5n + 6. \]

Rearranging this leads to:

\[ n^2 + 5n + 6 < 74, \]

which simplifies to:

\[ n^2 + 5n - 68 < 0. \]

### Step 5: Solve the quadratic inequality

To find the roots of the quadratic equation \(n^2 + 5n - 68 = 0\), we can use the quadratic formula:

\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-68)}}{2 \cdot 1} = \frac{-5 \pm \sqrt{25 + 272}}{2} = \frac{-5 \pm \sqrt{297}}{2}. \]

Calculating \(\sqrt{297} \approx 17.23\):

\[ n \approx \frac{-5 + 17.23}{2} \approx 6.115 \quad \text{and} \quad n \approx \frac{-5 - 17.23}{2} \text{ (not relevant as } n \text{ must be positive)}. \]

### Step 6: Determine the largest integer \(n\)

Since \(n\) must be a natural number, we check the integer values below \(6.115\):

- For \(n = 6\):
\[ 6^2 + 5 \cdot 6 - 68 = 36 + 30 - 68 = -2 < 0 \quad \text{(valid)} \]

- For \(n = 7\):
\[ 7^2 + 5 \cdot 7 - 68 = 49 + 35 - 68 = 16 > 0 \quad \text{(not valid)} \]

Thus, the largest value of \(n\) in \(\mathbb{N}\) for which the inequality holds is:

\[ \boxed{6}. \]