If `N` is a prime number which divides `S=^(39)P_(19)+^(38)P_(19)+^(37)P_(19)+…+"^(20)P_(19)`, then the largest possible value of `N` among following is

Options

A. `41`
B. `31`
C. `37`
D. `19`

Explanation

To solve the problem, we need to evaluate the expression \( S = {}^{39}P_{19} + {}^{38}P_{19} + {}^{37}P_{19} + \ldots + {}^{20}P_{19} \) and find the largest prime number \( N \) that divides \( S \).

### Step-by-Step Solution:

1. **Understanding Permutations**:
The notation \( {}^{n}P_{r} \) represents the number of permutations of \( n \) items taken \( r \) at a time, which is given by the formula:
\[ {}^{n}P_{r} = \frac{n!}{(n-r)!} \]

2. **Expressing \( S \)**:
We can express \( S \) using the formula for permutations:
\[ S = {}^{39}P_{19} + {}^{38}P_{19} + {}^{37}P_{19} + \ldots + {}^{20}P_{19} \]
This can be rewritten as:
\[ S = \frac{39!}{(39-19)!} + \frac{38!}{(38-19)!} + \frac{37!}{(37-19)!} + \ldots + \frac{20!}{(20-19)!} \]

3. **Factoring Out Common Terms**:
Notice that each term in \( S \) has a common factor of \( 19! \):
\[ S = 19! \left( {}^{39}C_{19} + {}^{38}C_{19} + {}^{37}C_{19} + \ldots + {}^{20}C_{19} \right) \]

4. **Using the Hockey Stick Identity**:
The Hockey Stick Identity in combinatorics states that:
\[ {}^{n}C_{r} + {}^{n-1}C_{r} + {}^{n-2}C_{r} + \ldots + {}^{r}C_{r} = {}^{n+1}C_{r+1} \]
Applying this identity, we have:
\[ {}^{39}C_{19} + {}^{38}C_{19} + \ldots + {}^{20}C_{19} = {}^{40}C_{20} \]
Thus, we can rewrite \( S \) as:
\[ S = 19! \cdot {}^{40}C_{20} \]

5. **Finding \( {}^{40}C_{20} \)**:
The binomial coefficient \( {}^{40}C_{20} \) is given by:
\[ {}^{40}C_{20} = \frac{40!}{20! \cdot 20!} \]

6. **Finding the Prime Factorization**:
Now we need to find the largest prime \( N \) that divides \( S \). Since \( S = 19! \cdot {}^{40}C_{20} \), we need to consider the prime factors of both \( 19! \) and \( {}^{40}C_{20} \).

7. **Identifying the Largest Prime**:
The largest prime number less than or equal to 40 is 37. However, we also need to check if 41 divides \( S \):
- \( 41 \) does not divide \( 19! \) since \( 19! \) contains primes only up to 19.
- To check \( {}^{40}C_{20} \):
- \( 40! \) contains \( 41 \) as a factor since \( 41 \) is prime and greater than \( 40 \).
- Therefore, \( 41 \) divides \( {}^{40}C_{20} \).

8. **Conclusion**:
The largest prime \( N \) that divides \( S \) is:
\[ N = 41 \]

### Final Answer:
The largest possible value of \( N \) is \( 41 \).