Is it `Ture` or `False`
`1+""^(3)C_(1)+""^(4)C_(2)=""^(5)C_(3)`.

Answer

To determine whether the statement \( 1 + \binom{3}{1} + \binom{4}{2} = \binom{5}{3} \) is true or false, we will evaluate both sides of the equation step by step.

### Step 1: Calculate \( \binom{3}{1} \)
The binomial coefficient \( \binom{n}{r} \) is calculated using the formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
For \( \binom{3}{1} \):
\[ \binom{3}{1} = \frac{3!}{1!(3-1)!} = \frac{3!}{1! \cdot 2!} = \frac{3 \times 2 \times 1}{1 \times (2 \times 1)} = \frac{6}{2} = 3 \]

### Step 2: Calculate \( \binom{4}{2} \)
Now, we calculate \( \binom{4}{2} \):
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2! \cdot 2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1) \cdot (2 \times 1)} = \frac{24}{4} = 6 \]

### Step 3: Calculate the left-hand side (LHS)
Now we can compute the left-hand side of the equation:
\[ 1 + \binom{3}{1} + \binom{4}{2} = 1 + 3 + 6 = 10 \]

### Step 4: Calculate \( \binom{5}{3} \)
Next, we calculate \( \binom{5}{3} \):
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \cdot 2!} = \frac{5 \times 4 \times 3!}{3! \cdot (2 \times 1)} = \frac{20}{2} = 10 \]

### Step 5: Compare LHS and RHS
Now we compare both sides:
\[ LHS = 10 \quad \text{and} \quad RHS = 10 \]

Since both sides are equal, we conclude that:
\[ 1 + \binom{3}{1} + \binom{4}{2} = \binom{5}{3} \]

Thus, the statement is **True**.