Prove that for n ≥ 4, nC3 + nC4 = n+1C4.

Steps to Solve

1. Definitions of Binomial Coefficients

The binomial coefficient ${n \choose r}$ is defined as:

$${n \choose r} = \frac{n!}{r!(n - r)!}$$

2. Write the Left Hand Side

We need to calculate the left-hand side of the equation ( nC_3 + nC_4 ):

$$ nC_3 + nC_4 = {n \choose 3} + {n \choose 4} $$

Inserting the definitions:

$$ {n \choose 3} + {n \choose 4} = \frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} $$

3. Common Denominator

The common denominator for the two fractions is ( 4!(n-4)! ):

$$ {n \choose 3} + {n \choose 4} = \frac{n! \cdot 4}{4! \cdot (n - 3)!} + \frac{n!}{4! \cdot (n - 4)!} $$

4. Combine the Fractions

Now combine the fractions:

$$ = \frac{n! \cdot 4(n - 4) + n!}{4! \cdot (n - 4)!} $$

This simplifies to:

$$ = \frac{n!(4(n - 4) + 1)}{4! \cdot (n - 4)!} $$

5. Simplifying the Expression

Let’s simplify ( 4(n - 4) + 1 = 4n - 16 + 1 = 4n - 15 ):

$$ = \frac{n!(4n - 15)}{4!(n - 4)!} $$

6. Write the Right Hand Side

Now, let’s calculate the right-hand side ( {n+1 \choose 4} ):

$$ {n+1 \choose 4} = \frac{(n + 1)!}{4!(n - 3)!} $$

7. Rewrite Right Hand Side

We can express ( (n + 1)! ) as ( (n + 1) n! ):

$$ = \frac{(n + 1)n!}{4!(n - 3)!} $$

8. Prove Equality

Now we want to show:

$$ \frac{n!(4n - 15)}{4!(n - 4)!} = \frac{(n + 1)n!}{4!(n - 3)!} $$

This simplifies to showing:

$$ 4n - 15 = 4(n + 1 - 4) $$

Checking:

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

9. Solve the last equation

Setting:

$$ 4n - 15 = 4n - 12 $$

This holds true, confirming that both sides of the original equation equal.