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
We start by rewriting the binomial coefficients in terms of factorials:

\[ \binom{4}{n} = \frac{4!}{n!(4-n)!}, \quad \binom{5}{n} = \frac{5!}{n!(5-n)!}, \quad \binom{6}{n} = \frac{6!}{n!(6-n)!} \]

### Step 2: Substitute into the Equation
Substituting these into the equation gives us:

\[ \frac{1}{\frac{4!}{n!(4-n)!}} = \frac{1}{\frac{5!}{n!(5-n)!}} + \frac{1}{\frac{6!}{n!(6-n)!}} \]

This simplifies to:

\[ \frac{n!(4-n)!}{4!} = \frac{n!(5-n)!}{5!} + \frac{n!(6-n)!}{6!} \]

### Step 3: Simplify the Equation
Cancelling \(n!\) from both sides (assuming \(n! \neq 0\)) gives:

\[ \frac{(4-n)!}{4!} = \frac{(5-n)!}{5!} + \frac{(6-n)!}{6!} \]

### Step 4: Express Factorials
Now we can express the factorials:

\[ \frac{(4-n)!}{4!} = \frac{(5-n)(4-n)!}{5 \cdot 4!} + \frac{(6-n)(5-n)(4-n)!}{6 \cdot 5 \cdot 4!} \]

### Step 5: Multiply Through by \(4! \cdot 30\)
To eliminate the denominators, multiply through by \(30\):

\[ 30(4-n)! = 6(5-n)(4-n)! + 5(6-n)(5-n)(4-n)! \]

### Step 6: Cancel \((4-n)!\)
Assuming \(4-n \neq 0\) (which implies \(n \neq 4\)), we can cancel \((4-n)!\):

\[ 30 = 6(5-n) + 5(6-n)(5-n) \]

### Step 7: Expand and Rearrange
Expanding gives:

\[ 30 = 30 - 6n + 5(30 - 11n + n^2) \]

This simplifies to:

\[ 30 = 30 - 6n + 150 - 55n + 5n^2 \]

Combining like terms yields:

\[ 0 = 5n^2 - 61n + 150 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ n = \frac{61 \pm \sqrt{61^2 - 4 \cdot 5 \cdot 150}}{2 \cdot 5} \]

Calculating the discriminant:

\[ 61^2 - 3000 = 3721 - 3000 = 721 \]

Thus:

\[ n = \frac{61 \pm 27}{10} \]

Calculating the two possible values:

1. \(n = \frac{88}{10} = 8.8\) (not valid since \(n\) must be an integer)
2. \(n = \frac{34}{10} = 3.4\) (not valid since \(n\) must be an integer)

### Step 9: Check Integer Values
Since \(n\) must be less than 4, we check \(n = 2\) and \(n = 3\):

- For \(n = 2\):
\[ \frac{1}{\binom{4}{2}} = \frac{1}{6}, \quad \frac{1}{\binom{5}{2}} + \frac{1}{\binom{6}{2}} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} \]
This holds true.

- For \(n = 3\):
\[ \frac{1}{\binom{4}{3}} = 1, \quad \frac{1}{\binom{5}{3}} + \frac{1}{\binom{6}{3}} = \frac{1}{10} + \frac{1}{20} = \frac{2 + 1}{20} = \frac{3}{20} \neq 1 \]

Thus, the only valid solution is:

\[ \boxed{2} \]