`""sum_(r=1)^(4)""^(21-r)C_(4)+ 17C_(5), is `

Options

A. `""^(21)C_(6)`
B. `""46C_(8)`
C. `""^(21)C_(7)`
D. `""^(21)C_(5)`

Answer

To solve the problem, we need to evaluate the expression:

\[ \sum_{r=1}^{4} \binom{21-r}{4} + \binom{17}{5} \]

### Step 1: Expand the Summation

We will first expand the summation by substituting the values of \( r \) from 1 to 4.

- For \( r = 1 \):
\[ \binom{21-1}{4} = \binom{20}{4} \]

- For \( r = 2 \):
\[ \binom{21-2}{4} = \binom{19}{4} \]

- For \( r = 3 \):
\[ \binom{21-3}{4} = \binom{18}{4} \]

- For \( r = 4 \):
\[ \binom{21-4}{4} = \binom{17}{4} \]

Thus, we can rewrite the summation as:
\[ \binom{20}{4} + \binom{19}{4} + \binom{18}{4} + \binom{17}{4} \]

### Step 2: Add \(\binom{17}{5}\)

Now we add \(\binom{17}{5}\) to the summation:
\[ \binom{20}{4} + \binom{19}{4} + \binom{18}{4} + \binom{17}{4} + \binom{17}{5} \]

### Step 3: Apply the Identity

We can use the identity:
\[ \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \]

We will apply this identity to combine \(\binom{17}{4}\) and \(\binom{17}{5}\):
\[ \binom{17}{4} + \binom{17}{5} = \binom{18}{5} \]

### Step 4: Combine Terms

Now we rewrite the expression:
\[ \binom{20}{4} + \binom{19}{4} + \binom{18}{4} + \binom{18}{5} \]

Next, we apply the identity again:
- Combine \(\binom{18}{4}\) and \(\binom{18}{5}\):
\[ \binom{18}{4} + \binom{18}{5} = \binom{19}{5} \]

Now we have:
\[ \binom{20}{4} + \binom{19}{4} + \binom{19}{5} \]

### Step 5: Apply the Identity Again

Now we can combine \(\binom{19}{4}\) and \(\binom{19}{5}\):
\[ \binom{19}{4} + \binom{19}{5} = \binom{20}{5} \]

So we can rewrite the expression as:
\[ \binom{20}{4} + \binom{20}{5} \]

### Step 6: Final Application of the Identity

Finally, we apply the identity one last time:
\[ \binom{20}{4} + \binom{20}{5} = \binom{21}{5} \]

### Conclusion

Thus, the final answer is:
\[ \binom{21}{5} \]