The number `N=""^(20)C_(7)-""^(20)C_(8)+""^(20)C_(9)-""^(20)C_(10)+….. -""^(20)C_(20)` is not divisible by :

Options

A. A) 3
B. B) 7
C. C) 11
D. D) 19

Answer

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

\[ N = \binom{20}{7} - \binom{20}{8} + \binom{20}{9} - \binom{20}{10} + \ldots - \binom{20}{20} \]

### Step 1: Rewrite the expression using summation notation
We can express \(N\) in summation notation as follows:

\[ N = \sum_{r=7}^{20} (-1)^{r-7} \binom{20}{r} \]

### Step 2: Use the Binomial Theorem
We can utilize the Binomial Theorem, which states that:

\[ (1 + x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r \]

For our case, we can consider \(x = -1\):

\[ (1 - 1)^{20} = \sum_{r=0}^{20} \binom{20}{r} (-1)^r = 0 \]

### Step 3: Split the summation
We can split the summation into two parts: one for even \(r\) and one for odd \(r\):

\[ 0 = \sum_{r=0}^{20} \binom{20}{r} (-1)^r = \sum_{r \text{ even}} \binom{20}{r} - \sum_{r \text{ odd}} \binom{20}{r} \]

### Step 4: Relate to our expression
The expression \(N\) can be related to the above summation. We can rewrite \(N\) as:

\[ N = \sum_{r=0}^{20} (-1)^r \binom{20}{r} - \left( \binom{20}{0} + \binom{20}{1} + \binom{20}{2} + \ldots + \binom{20}{6} \right) \]

### Step 5: Calculate the first part
From the Binomial Theorem, we know:

\[ \sum_{r=0}^{20} (-1)^r \binom{20}{r} = 0 \]

Thus, we have:

\[ N = -\left( \binom{20}{0} + \binom{20}{1} + \binom{20}{2} + \ldots + \binom{20}{6} \right) \]

### Step 6: Calculate the binomial coefficients
Now we need to calculate the sum of the first 7 binomial coefficients:

\[ \binom{20}{0} = 1, \quad \binom{20}{1} = 20, \quad \binom{20}{2} = 190, \quad \binom{20}{3} = 1140, \quad \binom{20}{4} = 4845, \quad \binom{20}{5} = 15504, \quad \binom{20}{6} = 38760 \]

Calculating the total:

\[ 1 + 20 + 190 + 1140 + 4845 + 15504 + 38760 = 60420 \]

### Step 7: Substitute back into \(N\)
Thus, we have:

\[ N = -60420 \]

### Step 8: Prime factorization of \(N\)
Now we perform the prime factorization of \(60420\):

\[ 60420 = 2^2 \times 3 \times 5 \times 7 \times 17 \]

### Step 9: Determine divisibility
Now we check which of the options \(3, 7, 11, 19\) \(N\) is not divisible by:

- \(N\) is divisible by \(3\)
- \(N\) is divisible by \(7\)
- \(N\) is divisible by \(19\)
- \(N\) is **not divisible by \(11\)**

### Final Answer
Thus, the number \(N\) is not divisible by:

\[ \text{Option C: } 11 \]