$\text{ }$
[asy]
unitsize(11);
for(int i=0; i<6; ++i)
{
    if(i<5)
        draw( (i, 0)--(i,5) );
    else draw( (i, 0)--(i,2) );

    if(i < 3)
        draw((0,i)--(5,i));
    else draw((0,i)--(4,i));
}
[/asy]
We are dividing the above figure into parts  with shapes: [asy]

unitsize(11);
draw((0,0)--(0,2));
draw((1,0)--(1,2));
draw((2,1)--(2,2));
draw((0,0)--(1,0));
draw((0,1)--(2,1));
draw((0,2)--(2,2));
[/asy][asy]
unitsize(11);
draw((0,0)--(0,2));
draw((1,0)--(1,2));
draw((2,1)--(2,2));
draw((3,1)--(3,2));
draw((0,0)--(1,0));
draw((0,1)--(3,1));
draw((0,2)--(3,2));
[/asy]
After that division, find the number of 
[asy]

unitsize(11);
draw((0,0)--(0,2));
draw((1,0)--(1,2));
draw((2,1)--(2,2));
draw((0,0)--(1,0));
draw((0,1)--(2,1));
draw((0,2)--(2,2));
[/asy]
shaped parts.
1. **Identify the shapes and their counts:**
   - Let \( x \) be the number of shapes with three cells.
   - Let \( y \) be the number of shapes with four cells.
   - The total number of cells in the figure is \( 22 \).

2. **Set up the equation:**
   - Each shape with three cells contributes 3 cells.
   - Each shape with four cells contributes 4 cells.
   - Therefore, the equation is:
     \[
     3x + 4y = 22
     \]

3. **Solve the equation for integer solutions:**
   - We need to find non-negative integer solutions for \( x \) and \( y \).
   - By trial and error or systematic solving, we find:
     \[
     (x, y) = (6, 1) \quad \text{or} \quad (x, y) = (2, 4)
     \]

4. **Verify the solutions:**
   - For \( (x, y) = (6, 1) \):
     \[
     3(6) + 4(1) = 18 + 4 = 22
     \]
     This solution is valid.
   - For \( (x, y) = (2, 4) \):
     \[
     3(2) + 4(4) = 6 + 16 = 22
     \]
     This solution is also valid.

5. **Determine the correct configuration:**
   - We need to check if the configurations can be drawn correctly.
   - The problem suggests that the configuration with \( (x, y) = (6, 1) \) can be drawn as shown in the provided diagram.

6. **Conclusion:**
   - The number of shapes with three cells is \( 6 \).

The final answer is \( \boxed{ 6 } \).