If `(1)/(""^(4)C_(n))=(1)/(""^(5)C_(n))+(1)/(""^(6)C_(n))`, then value of n is:

Options

A. 3
B. 4
C. 1
D. 2

Answer

To solve the equation \(\frac{1}{\binom{4}{n}} = \frac{1}{\binom{5}{n}} + \frac{1}{\binom{6}{n}}\), we will follow these steps:

### Step 1: Rewrite the Binomial Coefficients
Recall the formula for binomial coefficients:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Using this, we can express the equation as:
\[ \frac{1}{\binom{4}{n}} = \frac{n!(4-n)!}{4!} \]
\[ \frac{1}{\binom{5}{n}} = \frac{n!(5-n)!}{5!} \]
\[ \frac{1}{\binom{6}{n}} = \frac{n!(6-n)!}{6!} \]

### Step 2: Substitute into the Equation
Substituting these into the original equation gives us:
\[ \frac{n!(4-n)!}{4!} = \frac{n!(5-n)!}{5!} + \frac{n!(6-n)!}{6!} \]

### Step 3: Cancel \(n!\)
Since \(n!\) is common in all terms, we can cancel it out (assuming \(n! \neq 0\)):
\[ \frac{(4-n)!}{4!} = \frac{(5-n)!}{5!} + \frac{(6-n)!}{6!} \]

### Step 4: Rewrite Factorials
We can rewrite \(5!\) and \(6!\) in terms of \(4!\):
\[ \frac{(4-n)!}{4!} = \frac{(5-n)(4-n)!}{5 \cdot 4!} + \frac{(6-n)(5-n)(4-n)!}{6 \cdot 5 \cdot 4!} \]
Cancelling \((4-n)!\) gives:
\[ \frac{1}{4!} = \frac{5-n}{5 \cdot 4!} + \frac{(6-n)(5-n)}{6 \cdot 5 \cdot 4!} \]

### Step 5: Multiply through by \(4! \cdot 30\) (LCM)
Multiply through by \(30\) to eliminate the denominators:
\[ 30 = 6(5-n) + (6-n)(5-n) \]

### Step 6: Expand and Simplify
Expanding the right side:
\[ 30 = 30 - 6n + 30 - 11n + n^2 \]
Combine like terms:
\[ 0 = n^2 - 17n + 30 \]

### Step 7: Factor the Quadratic
Factoring gives:
\[ (n - 15)(n - 2) = 0 \]
Thus, \(n = 15\) or \(n = 2\).

### Step 8: Check Validity
Since \(n\) must be less than or equal to \(4\) (as \(n\) is the lower index in the binomial coefficient), we discard \(n = 15\) and accept \(n = 2\).

### Final Answer
Thus, the value of \(n\) is:
\[ \boxed{2} \]