Prove that :
`""^(2)C_(1)+ ""^(3)C_(1)+""^(4)C_(1)=""^(3)C_(2)+""^(4)C_(2)`.

Answer

To prove that \( \binom{2}{1} + \binom{3}{1} + \binom{4}{1} = \binom{3}{2} + \binom{4}{2} \), we will evaluate both sides of the equation.

### Step 1: Evaluate the Left Hand Side (LHS)

The left-hand side is given by:

\[ \binom{2}{1} + \binom{3}{1} + \binom{4}{1} \]

Using the property of combinations, we know that \( \binom{n}{1} = n \). Thus, we can calculate each term:

\[ \binom{2}{1} = 2, \quad \binom{3}{1} = 3, \quad \binom{4}{1} = 4 \]

Now, substituting these values into the LHS:

\[ LHS = 2 + 3 + 4 = 9 \]

### Step 2: Evaluate the Right Hand Side (RHS)

The right-hand side is given by:

\[ \binom{3}{2} + \binom{4}{2} \]

Again, using the formula for combinations, we can calculate each term:

\[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2! \cdot 1!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3 \]

\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2! \cdot 2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{12}{4} = 6 \]

Now, substituting these values into the RHS:

\[ RHS = 3 + 6 = 9 \]

### Step 3: Compare LHS and RHS

Now we compare both sides:

\[ LHS = 9 \quad \text{and} \quad RHS = 9 \]

Since both sides are equal, we have:

\[ \binom{2}{1} + \binom{3}{1} + \binom{4}{1} = \binom{3}{2} + \binom{4}{2} \]

Thus, we have proved the statement.