For n = 6, let
`N=(""^(n)C_(0))^(2)+(""^(n)C_(1))^(2)+…+(""^(n)C_(n))^(2)`
If N- 204 = m!, then m = ___________

Answer

To solve the problem, we need to calculate the value of \( N \) for \( n = 6 \) and then find \( m \) such that \( N - 204 = m! \).

### Step-by-Step Solution:

1. **Understanding the Expression for \( N \)**:
The expression for \( N \) is given as:
\[ N = \sum_{r=0}^{n} \binom{n}{r}^2 \]
For \( n = 6 \), this becomes:
\[ N = \binom{6}{0}^2 + \binom{6}{1}^2 + \binom{6}{2}^2 + \binom{6}{3}^2 + \binom{6}{4}^2 + \binom{6}{5}^2 + \binom{6}{6}^2 \]

2. **Calculating Each Binomial Coefficient**:
We calculate the binomial coefficients:
- \( \binom{6}{0} = 1 \)
- \( \binom{6}{1} = 6 \)
- \( \binom{6}{2} = 15 \)
- \( \binom{6}{3} = 20 \)
- \( \binom{6}{4} = 15 \)
- \( \binom{6}{5} = 6 \)
- \( \binom{6}{6} = 1 \)

3. **Calculating \( N \)**:
Now we square each coefficient and sum them up:
\[ N = 1^2 + 6^2 + 15^2 + 20^2 + 15^2 + 6^2 + 1^2 \]
\[ N = 1 + 36 + 225 + 400 + 225 + 36 + 1 \]
\[ N = 1 + 36 + 225 + 400 + 225 + 36 + 1 = 924 \]

4. **Finding \( m \)**:
We know from the problem that:
\[ N - 204 = m! \]
Substituting \( N \):
\[ 924 - 204 = m! \]
\[ 720 = m! \]

5. **Determining \( m \)**:
We need to find \( m \) such that \( m! = 720 \). Checking factorials:
- \( 1! = 1 \)
- \( 2! = 2 \)
- \( 3! = 6 \)
- \( 4! = 24 \)
- \( 5! = 120 \)
- \( 6! = 720 \)

Thus, \( m = 6 \).

### Final Answer:
The value of \( m \) is \( \boxed{6} \).