If `""^(n-1) C_6+""^(n-1) C_7 gt ""^n C_6` then n is greater than

Options

A. 4
B. 12
C. 13
D. 15

Answer

To solve the inequality \( \binom{n-1}{6} + \binom{n-1}{7} > \binom{n}{6} \), we can use the properties of combinations and some algebraic manipulation.

### Step-by-Step Solution:

1. **Apply the Combination Identity**:
We know from the properties of combinations that:
\[ \binom{n-1}{r} + \binom{n-1}{r+1} = \binom{n}{r+1} \]
In our case, we can set \( r = 6 \):
\[ \binom{n-1}{6} + \binom{n-1}{7} = \binom{n}{7} \]
Therefore, we can rewrite the original inequality as:
\[ \binom{n}{7} > \binom{n}{6} \]

2. **Use the Definition of Combinations**:
The combinations can be expressed as:
\[ \binom{n}{7} = \frac{n!}{7!(n-7)!} \quad \text{and} \quad \binom{n}{6} = \frac{n!}{6!(n-6)!} \]
Substituting these into the inequality gives:
\[ \frac{n!}{7!(n-7)!} > \frac{n!}{6!(n-6)!} \]

3. **Cancel Out Common Terms**:
Since \( n! \) is common on both sides, we can cancel it out (assuming \( n! \neq 0 \)):
\[ \frac{1}{7!(n-7)!} > \frac{1}{6!(n-6)!} \]

4. **Cross-Multiply**:
Cross-multiplying gives:
\[ 6!(n-6)! > 7!(n-7)! \]

5. **Simplify the Factorials**:
We can simplify \( 7! = 7 \times 6! \):
\[ 6!(n-6)! > 7 \times 6!(n-7)! \]
Dividing both sides by \( 6! \) (assuming \( 6! \neq 0 \)):
\[ (n-6)! > 7(n-7)! \]

6. **Further Simplification**:
We can express \( (n-6)! \) as \( (n-6)(n-7)! \):
\[ (n-6)(n-7)! > 7(n-7)! \]
Dividing both sides by \( (n-7)! \) (assuming \( (n-7)! \neq 0 \)):
\[ n-6 > 7 \]

7. **Solve for \( n \)**:
Rearranging gives:
\[ n > 13 \]

### Conclusion:
Thus, the solution to the inequality \( \binom{n-1}{6} + \binom{n-1}{7} > \binom{n}{6} \) is:
\[ n > 13 \]