`""^(n)C_(n-r)+3.""^(n)C_(n-r+1)+3.""^(n)C_(n-r+2)+""^(n)C_(n-r+3)=""^(x)C_(r)`

Explanation

To solve the equation

\[ \binom{n}{n - r + 3} + 3 \cdot \binom{n}{n - r + 1} + 3 \cdot \binom{n}{n - r + 2} + \binom{n}{n - r + 3} = \binom{x}{r}, \]

we will simplify the left-hand side using properties of combinations.

### Step 1: Rewrite the combinations
First, we can rewrite the combinations using the property \(\binom{n}{k} = \binom{n}{n-k}\):

\[ \binom{n}{n - r + 3} = \binom{n}{r - 3}, \quad \binom{n}{n - r + 1} = \binom{n}{r - 1}, \quad \binom{n}{n - r + 2} = \binom{n}{r - 2}. \]

Thus, we can rewrite the equation as:

\[ \binom{n}{r - 3} + 3 \cdot \binom{n}{r - 1} + 3 \cdot \binom{n}{r - 2} + \binom{n}{r - 3} = \binom{x}{r}. \]

### Step 2: Combine like terms
Now, combine the terms:

\[ 2 \cdot \binom{n}{r - 3} + 3 \cdot \binom{n}{r - 1} + 3 \cdot \binom{n}{r - 2} = \binom{x}{r}. \]

### Step 3: Use the Hockey Stick Identity
We can apply the Hockey Stick Identity, which states:

\[ \binom{n}{k} + \binom{n}{k+1} + \ldots + \binom{n}{m} = \binom{n+1}{m+1}, \]

to simplify the left side.

1. We first consider \(2 \cdot \binom{n}{r - 3}\) as \(\binom{n}{r - 3} + \binom{n}{r - 3}\).
2. Then we can group the terms as follows:

\[ \binom{n}{r - 3} + \binom{n}{r - 2} + \binom{n}{r - 1} + \binom{n}{r - 1} + \binom{n}{r - 2} + \binom{n}{r - 3} = \binom{n+1}{r - 2} + \binom{n+1}{r - 1} + \binom{n+1}{r - 3}. \]

### Step 4: Apply Hockey Stick Identity
Using the Hockey Stick Identity again, we have:

\[ \binom{n+1}{r - 2} + \binom{n+1}{r - 1} + \binom{n+1}{r - 3} = \binom{n + 3}{r}. \]

### Step 5: Equate to \(\binom{x}{r}\)
Now we can equate this to the right-hand side of the original equation:

\[ \binom{n + 3}{r} = \binom{x}{r}. \]

### Step 6: Solve for \(x\)
Since the binomial coefficients are equal, we can conclude:

\[ x = n + 3. \]

### Final Answer
Thus, the value of \(x\) is:

\[ \boxed{n + 3}. \]