If `""^(n-1)C_4-""^(n-1)C_3-5/4"^(n-2)P_2 lt 0` then the number of values of `n in N` has is

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the inequality \( \binom{n-1}{4} - \binom{n-1}{3} - \frac{5}{4} P(n-2, 2) < 0 \), we will follow these steps:

### Step 1: Rewrite the Binomial and Permutation Expressions
Using the formulas for combinations and permutations, we can express the terms in the inequality:
- \( \binom{n-1}{4} = \frac{(n-1)(n-2)(n-3)(n-4)}{4!} = \frac{(n-1)(n-2)(n-3)(n-4)}{24} \)
- \( \binom{n-1}{3} = \frac{(n-1)(n-2)(n-3)}{3!} = \frac{(n-1)(n-2)(n-3)}{6} \)
- \( P(n-2, 2) = \frac{(n-2)!}{(n-4)!} = (n-2)(n-3) \)

Substituting these into the inequality gives:
\[ \frac{(n-1)(n-2)(n-3)(n-4)}{24} - \frac{(n-1)(n-2)(n-3)}{6} - \frac{5}{4} \cdot (n-2)(n-3) < 0 \]

### Step 2: Simplify the Inequality
To simplify, multiply the entire inequality by 24 (to eliminate the denominators):
\[ (n-1)(n-2)(n-3)(n-4) - 4(n-1)(n-2)(n-3) - 30(n-2)(n-3) < 0 \]

### Step 3: Factor Out Common Terms
Notice that \( (n-2)(n-3) \) is a common factor:
\[ (n-2)(n-3) \left( (n-1)(n-4) - 4(n-1) - 30 \right) < 0 \]

### Step 4: Expand and Simplify Further
Expanding the expression inside the parentheses:
\[ (n-1)(n-4) - 4(n-1) - 30 = n^2 - 5n + 4 - 4n + 4 - 30 = n^2 - 9n - 22 \]

So the inequality becomes:
\[ (n-2)(n-3)(n^2 - 9n - 22) < 0 \]

### Step 5: Factor the Quadratic
Now we need to factor \( n^2 - 9n - 22 \):
The roots can be found using the quadratic formula:
\[ n = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot (-22)}}{2 \cdot 1} = \frac{9 \pm \sqrt{81 + 88}}{2} = \frac{9 \pm \sqrt{169}}{2} = \frac{9 \pm 13}{2} \]
This gives us the roots \( n = 11 \) and \( n = -2 \).

Thus, we can factor the quadratic as:
\[ (n - 11)(n + 2) \]

### Step 6: Combine Factors
Now we have:
\[ (n-2)(n-3)(n-11)(n+2) < 0 \]

### Step 7: Identify Critical Points
The critical points are \( n = -2, 2, 3, 11 \). We will analyze the sign of the product in the intervals determined by these points.

### Step 8: Test Intervals
1. **Interval \( (-\infty, -2) \)**: All factors negative → Positive
2. **Interval \( (-2, 2) \)**: \( (n+2) \) positive, others negative → Negative
3. **Interval \( (2, 3) \)**: \( (n-2) \) positive, others negative → Positive
4. **Interval \( (3, 11) \)**: All positive except \( (n-11) \) → Negative
5. **Interval \( (11, \infty) \)**: All factors positive → Positive

### Step 9: Determine Valid Intervals
The inequality is satisfied in the intervals \( (-2, 2) \) and \( (3, 11) \).

### Step 10: Find Natural Number Solutions
The natural numbers in these intervals are:
- From \( (3, 11) \): \( 4, 5, 6, 7, 8, 9, 10 \) (7 values)
- From \( (-2, 2) \): No natural numbers.

### Conclusion
Thus, the total number of values of \( n \) in \( \mathbb{N} \) that satisfy the inequality is **7**.