Let `X=(\ ^(10)C_1)^2+2(\ ^(10)C_2)^2+3(\ ^(10)C_3)^2+\ ddot\ +10(\ ^(10)C_(10))^2` , where `\ ^(10)C_r` , `r in {1,\ 2,\ ddot,\ 10}` denote binomial coefficients. Then, the value of `1/(1430)\ X` is _________.

Explanation

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

\[ X = \sum_{r=1}^{10} r \cdot \binom{10}{r}^2 \]

### Step 1: Use the Identity for Binomial Coefficients
We can use the identity that relates the sum of products of binomial coefficients:

\[ \sum_{r=0}^{n} r \cdot \binom{n}{r} = n \cdot 2^{n-1} \]

This means we can express our sum in terms of binomial coefficients.

### Step 2: Rewrite the Expression
We can rewrite \(X\) as follows:

\[ X = \sum_{r=1}^{10} r \cdot \binom{10}{r}^2 = \sum_{r=1}^{10} r \cdot \binom{10}{r} \cdot \binom{10}{r} \]

### Step 3: Apply the Identity
Using the identity mentioned earlier, we can find that:

\[ \sum_{r=1}^{10} r \cdot \binom{10}{r}^2 = 10 \cdot \sum_{r=1}^{10} \binom{10-1}{r-1} \cdot \binom{10}{r} \]

### Step 4: Use Vandermonde's Identity
By applying Vandermonde's identity, we can simplify:

\[ \sum_{r=0}^{n} \binom{m}{r} \cdot \binom{n}{k-r} = \binom{m+n}{k} \]

In our case, we have \(m = n = 10\) and \(k = 10\):

\[ \sum_{r=0}^{10} \binom{10}{r} \cdot \binom{10}{10-r} = \binom{20}{10} \]

### Step 5: Calculate \(X\)
Thus, we can express \(X\) as:

\[ X = 10 \cdot \binom{20}{10} \]

### Step 6: Calculate \(\binom{20}{10}\)
Now we need to calculate \(\binom{20}{10}\):

\[ \binom{20}{10} = \frac{20!}{10! \cdot 10!} = 184756 \]

### Step 7: Substitute Back into \(X\)
Now substituting back into our expression for \(X\):

\[ X = 10 \cdot 184756 = 1847560 \]

### Step 8: Find \(\frac{X}{1430}\)
Finally, we need to compute:

\[ \frac{X}{1430} = \frac{1847560}{1430} \]

Calculating this gives:

\[ \frac{1847560}{1430} = 12920 \]

### Final Answer
Thus, the value of \(\frac{X}{1430}\) is:

\[ \boxed{12920} \]