find up to two decimal places the positive value of n if nc2=c(n,3)+c(n,4)

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.

Find up to two decimal places the positive value of n if nc2 = c(n,3) + c(n,4).

Discover how to solve the combinatorics equation \( \binom{n}{2} = \binom{n}{3} + \binom{n}{4} \) by expanding binomial coefficients and applying the quadratic formula. This detailed step-by-step solution is ideal for students in Grades 10-12.

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