STATEMENT -1 : There are 12 points in a plane of which only 5 are collinear , then the number of straight lines obained by joining these points in pairs is `""^(12)C_(2) - ""^(5)C_(2)` .
STATEMENT-2: `""^(n +1)C_(r) - ""^(n-1)C_(r - 1) = ""^(n)C_(r) + ""^(n)C_(r - 2)`
STATEMENT -3 :2n persons may be seated at two round tables , n person seated at each , in `((2n)!)/(n^(2))` in differnet ways.

Options

A. F T F
B. T T F
C. F F T
D. F T T

Answer

To solve the problem, we will analyze each statement one by one and determine their validity.

### Statement 1:
**Statement:** There are 12 points in a plane of which only 5 are collinear, then the number of straight lines obtained by joining these points in pairs is \( \binom{12}{2} - \binom{5}{2} \).

**Solution Steps:**
1. **Calculate Total Lines from All Points:**
The total number of ways to choose 2 points from 12 points is given by the combination formula:
\[ \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 \]

2. **Calculate Lines from Collinear Points:**
The number of lines that can be formed by choosing 2 points from the 5 collinear points is:
\[ \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \]

3. **Subtract the Collinear Lines:**
The number of lines that can be formed from the remaining points (not collinear) is:
\[ \text{Total Lines} - \text{Collinear Lines} = 66 - 10 = 56 \]

4. **Consider the Extra Line from Collinear Points:**
Since all 5 collinear points lie on the same line, we need to add back 1 for the line that passes through all 5 points:
\[ 56 + 1 = 57 \]

5. **Conclusion:**
Therefore, the statement that the number of lines is \( \binom{12}{2} - \binom{5}{2} \) is incorrect as it does not account for the extra line formed by the collinear points.

### Statement 2:
**Statement:** \( \binom{n + 1}{r} - \binom{n - 1}{r - 1} = \binom{n}{r} + \binom{n}{r - 2} \).

**Solution Steps:**
1. **Use Pascal's Identity:**
We know from combinatorial identities that:
\[ \binom{n + 1}{r} = \binom{n}{r} + \binom{n}{r - 1} \]
and
\[ \binom{n - 1}{r - 1} = \binom{n}{r - 1} - \binom{n - 1}{r - 2} \]

2. **Substituting in the Statement:**
Substituting the identities into the statement gives:
\[ \binom{n + 1}{r} - \left( \binom{n}{r - 1} - \binom{n - 1}{r - 2} \right) = \binom{n}{r} + \binom{n}{r - 2} \]

3. **Rearranging:**
This leads to:
\[ \binom{n + 1}{r} - \binom{n}{r - 1} + \binom{n - 1}{r - 2} = \binom{n}{r} + \binom{n}{r - 2} \]

4. **Conclusion:**
The statement is incorrect because the left-hand side does not equal the right-hand side as per the identities.

### Statement 3:
**Statement:** 2n persons may be seated at two round tables, n persons seated at each, in \( \frac{(2n)!}{n^2} \) different ways.

**Solution Steps:**
1. **Select n Persons from 2n:**
The number of ways to choose n persons from 2n is:
\[ \binom{2n}{n} \]

2. **Arrange n Persons in a Round Table:**
The arrangement of n persons in a round table is given by:
\[ (n-1)! \]
(since arrangements in a circle are counted by fixing one person).

3. **Repeat for the Second Table:**
The same applies for the remaining n persons at the second table:
\[ (n-1)! \]

4. **Combine the Counts:**
The total number of arrangements is:
\[ \binom{2n}{n} \times (n-1)! \times (n-1)! \]

5. **Simplifying:**
This can be expressed as:
\[ \frac{(2n)!}{n!n!} \times (n-1)! \times (n-1)! = \frac{(2n)!}{n!n!} \times \frac{(n!)^2}{n^2} = \frac{(2n)!}{n^2} \]

6. **Conclusion:**
Therefore, the statement is correct.

### Final Conclusion:
- **Statement 1:** False
- **Statement 2:** False
- **Statement 3:** True