STATEMENT -1 : No. of ways of distributing 20 different things equally among 5 persons = `20!//4!^(5)`.
STATEMENT -2: No. of ways of distributing 20 different things equally among 5 persons = `20!//4!^(5).5!` .
STATEMENT-3 : No. of ways of distributing 20 different things in 5 groups such that in three group there are 6 things each and in remaining two groups there is 1 thing each = `(20!)/((6!)^(3))` .

Options

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

Answer

To determine the validity of the statements regarding the distribution of 20 different things among 5 persons and groups, we will analyze each statement step by step.

### Step-by-Step Solution:

**Statement 1:**
- We need to find the number of ways to distribute 20 different things equally among 5 persons.
- Since there are 20 different items and we want to distribute them equally, each person will receive \( \frac{20}{5} = 4 \) items.
- The formula for distributing \( mn \) objects into \( n \) groups such that each group gets \( m \) objects is given by:
\[ \frac{(mn)!}{(m!)^n} \]
- Here, \( m = 4 \) and \( n = 5 \). Therefore, we have:
\[ \text{Total ways} = \frac{20!}{(4!)^5} \]
- Thus, **Statement 1 is true**.

**Statement 2:**
- This statement claims that the number of ways to distribute 20 different things equally among 5 persons is:
\[ \frac{20!}{(4!)^5 \cdot 5!} \]
- The additional factor of \( 5! \) is incorrect because \( 5! \) is used when we need to account for the arrangements of the groups themselves, but in this case, the groups are indistinguishable since each person gets the same number of items.
- Therefore, **Statement 2 is false**.

**Statement 3:**
- We need to find the number of ways to distribute 20 different things into 5 groups such that 3 groups have 6 items each and the remaining 2 groups have 1 item each.
- Here, we have:
- 3 groups getting 6 items each: total items = \( 3 \times 6 = 18 \)
- 2 groups getting 1 item each: total items = \( 2 \times 1 = 2 \)
- The total number of items is \( 18 + 2 = 20 \), which is correct.
- The formula for this distribution is:
\[ \frac{20!}{(6!)^3 \cdot (1!)^2} \]
- Since \( 1! = 1 \), we can simplify this to:
\[ \frac{20!}{(6!)^3} \]
- Thus, **Statement 3 is true**.

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