We are tasked with finding the positive value of $n$ up to two decimal places for the equation:

$\binom{n}{2} = \binom{n}{3} + \binom{n}{4}$

Step 1: Expanding the binomial coefficients

We can express the binomial coefficients using their definitions:

$\binom{n}{2} = \frac{n(n-1)}{2}$ $\binom{n}{3} = \frac{n(n-1)(n-2)}{6}$ $\binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24}$

Step 2: Substituting into the equation

Now substitute these expressions into the given equation:

$\frac{n(n-1)}{2} = \frac{n(n-1)(n-2)}{6} + \frac{n(n-1)(n-2)(n-3)}{24}$

Step 3: Eliminate the fractions

To simplify the equation, multiply through by 24 to eliminate the denominators:

$12n(n-1) = 4n(n-1)(n-2) + n(n-1)(n-2)(n-3)$

Step 4: Simplify the equation

Factor $n(n-1)(n-2)$ on the right-hand side:

$12n(n-1) = n(n-1)(n-2) \left(4 + (n-3)\right)$ $12n(n-1) = n(n-1)(n-2)(n+1)$

Step 5: Divide both sides by $n(n-1)$ (assuming $n \neq 0$ and $n \neq 1$):

$12 = (n-2)(n+1)$

Step 6: Expand and solve for $n$

Expanding the right-hand side:

$12 = n^2 - n - 2$

Rearrange the equation:

$n^2 - n - 14 = 0$

Step 7: Solve the quadratic equation

Use the quadratic formula to solve for $n$:

$n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-14)}}{2(1)}$ $n = \frac{1 \pm \sqrt{1 + 56}}{2}$ $n = \frac{1 \pm \sqrt{57}}{2}$ [ n = \frac{1 \pm 7.