Evaluate
` C(19,17)+ C(19,18)`

Answer

To evaluate \( C(19, 17) + C(19, 18) \), we can use a property of combinations. The property states that:

\[ C(n, r) + C(n, r+1) = C(n+1, r+1) \]

In our case, we have \( n = 19 \) and \( r = 17 \). Therefore, we can rewrite the expression as follows:

\[ C(19, 17) + C(19, 18) = C(20, 18) \]

Now, we need to calculate \( C(20, 18) \). The formula for combinations is given by:

\[ C(n, r) = \frac{n!}{r!(n-r)!} \]

Substituting \( n = 20 \) and \( r = 18 \):

\[ C(20, 18) = \frac{20!}{18!(20-18)!} = \frac{20!}{18! \cdot 2!} \]

Next, we can simplify this expression. The factorial \( 20! \) can be expressed as:

\[ 20! = 20 \times 19 \times 18! \]

Thus, we can substitute this back into our equation:

\[ C(20, 18) = \frac{20 \times 19 \times 18!}{18! \cdot 2!} \]

Now, we can cancel \( 18! \) from the numerator and the denominator:

\[ C(20, 18) = \frac{20 \times 19}{2!} \]

Calculating \( 2! \):

\[ 2! = 2 \times 1 = 2 \]

So we have:

\[ C(20, 18) = \frac{20 \times 19}{2} = \frac{380}{2} = 190 \]

Thus, the final answer is:

\[ C(19, 17) + C(19, 18) = 190 \]