The expression
`""^n C_r+4 ""^n C_(r-1) +6 ""^n C_(r-2) +4 ""^n C_(r-3)+""^n C(r-4)=`

Options

A. `"^(n+4) C_r`
B. `2, "^(n+4) C_(r-1)`
C. `4. "^n C_r`
D. `11, "^n C_r`

Answer

To solve the expression

\[ {n \choose r} + 4 {n \choose r-1} + 6 {n \choose r-2} + 4 {n \choose r-3} + {n \choose r-4} \]

we can use the properties of combinations and Pascal's identity.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We have a sum of combinations with different coefficients. The coefficients are 1, 4, 6, 4, and 1, which resemble the binomial coefficients from the expansion of \((x + y)^4\).

2. **Using Binomial Coefficients**:
The coefficients can be recognized as:
\[ {4 \choose 0}, {4 \choose 1}, {4 \choose 2}, {4 \choose 3}, {4 \choose 4} \]
Thus, we can rewrite the expression as:
\[ \sum_{k=0}^{4} {4 \choose k} {n \choose r-k} \]

3. **Applying Vandermonde's Identity**:
According to Vandermonde's identity, we have:
\[ \sum_{k=0}^{r} {m \choose k} {n \choose r-k} = {m+n \choose r} \]
In our case, \(m = 4\) and \(n = n\). Therefore, we can apply this identity:
\[ \sum_{k=0}^{4} {4 \choose k} {n \choose r-k} = {n+4 \choose r} \]

4. **Final Result**:
Thus, the value of the original expression simplifies to:
\[ {n+4 \choose r} \]

### Conclusion:
The final answer to the expression is:
\[ {n+4 \choose r} \]