Let `f:A->A` be an invertible function where `A= {1,2,3,4,5,6}` The number of these functions in which at least three elements have self image is

Explanation

To solve the problem, we need to find the number of invertible functions \( f: A \to A \) where \( A = \{1, 2, 3, 4, 5, 6\} \) such that at least three elements have self-images.

### Step 1: Understand the problem
We need to count the number of permutations of the set \( A \) where at least three elements map to themselves. This means we will consider cases where exactly 3, 4, 5, or 6 elements are fixed points.

### Step 2: Define the cases
Let \( r \) be the number of elements that have self-images. We will consider the following cases:
- Case 1: \( r = 3 \)
- Case 2: \( r = 4 \)
- Case 3: \( r = 5 \)
- Case 4: \( r = 6 \)

### Step 3: Calculate for each case

**Case 1: \( r = 3 \)**
- Choose 3 elements from 6 to be fixed points: \( \binom{6}{3} \)
- The remaining 3 elements must be deranged (none of them should map to themselves). The number of derangements \( D(n) \) for \( n \) elements is given by:
\[ D(n) = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]
For \( n = 3 \):
\[ D(3) = 3! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}\right) = 6 \left(1 - 1 + 0.5 - \frac{1}{6}\right) = 6 \left(0.5 - \frac{1}{6}\right) = 6 \left(\frac{3}{6} - \frac{1}{6}\right) = 6 \cdot \frac{2}{6} = 2 \]
- Total for this case:
\[ \binom{6}{3} \cdot D(3) = 20 \cdot 2 = 40 \]

**Case 2: \( r = 4 \)**
- Choose 4 elements from 6 to be fixed points: \( \binom{6}{4} \)
- The remaining 2 elements must be deranged. For \( n = 2 \):
\[ D(2) = 2! \left(1 - \frac{1}{1!} + \frac{1}{2!}\right) = 2 \left(1 - 1 + 0.5\right) = 2 \cdot 0.5 = 1 \]
- Total for this case:
\[ \binom{6}{4} \cdot D(2) = 15 \cdot 1 = 15 \]

**Case 3: \( r = 5 \)**
- Choose 5 elements from 6 to be fixed points: \( \binom{6}{5} \)
- The remaining 1 element must be deranged, which is not possible (D(1) = 0).
- Total for this case:
\[ \binom{6}{5} \cdot D(1) = 6 \cdot 0 = 0 \]

**Case 4: \( r = 6 \)**
- All elements are fixed points: \( \binom{6}{6} = 1 \).
- Total for this case:
\[ 1 \]

### Step 4: Combine the results
Now, we sum the totals from all cases:
\[ \text{Total} = 40 + 15 + 0 + 1 = 56 \]

### Final Answer
The number of invertible functions where at least three elements have self-images is **56**.