← Use the conditions that force a value first, then list what is left · Cube Stacks and Their Views

Use the conditions that force a value first, then list what is left · 12 practice problems

6.G.A.47.G.A.3

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: 3

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 2 by 2 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 2 by 2 block.
The front view is 2, 3 cells high, left to right.
The side view is 3, 3 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

0 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 0

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

3

3 stacks are possible.

Answer: 3

4 · Reviewdoes it hold up?

Every one of the 3 stacks uses the same footprint and reaches 3 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 3 in every cell would mean checking 81 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.

Variant 2 easy answer: 5

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 2 by 2 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 2 by 2 block.
The front view is 4, 4 cells high, left to right.
The side view is 4, 3 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

0 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 0

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

5

5 stacks are possible.

Answer: 5

4 · Reviewdoes it hold up?

Every one of the 5 stacks uses the same footprint and reaches 4 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 4 in every cell would mean checking 256 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.

Variant 3 easy answer: 3

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 2 by 2 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 2 by 2 block.
The front view is 3, 4 cells high, left to right.
The side view is 4, 2 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

0 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 0

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

3

3 stacks are possible.

Answer: 3

4 · Reviewdoes it hold up?

Every one of the 3 stacks uses the same footprint and reaches 4 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 4 in every cell would mean checking 256 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.

Variant 4 medium answer: 7

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 3 by 2 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 3 by 2 block.
The front view is 3, 2 cells high, left to right.
The side view is 3, 3, 3 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

0 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 0

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

7

7 stacks are possible.

Answer: 7

4 · Reviewdoes it hold up?

Every one of the 7 stacks uses the same footprint and reaches 3 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 3 in every cell would mean checking 729 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.

Variant 5 medium answer: 3

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 3 by 2 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 3 by 2 block.
The front view is 3, 3 cells high, left to right.
The side view is 3, 2, 1 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

2 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 2

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

3

3 stacks are possible.

Answer: 3

4 · Reviewdoes it hold up?

Every one of the 3 stacks uses the same footprint and reaches 3 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 3 in every cell would mean checking 729 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.

Variant 6 medium answer: 7

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 2 by 3 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 2 by 3 block.
The front view is 1, 2, 2 cells high, left to right.
The side view is 2, 2 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

2 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 2

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

7

7 stacks are possible.

Answer: 7

4 · Reviewdoes it hold up?

Every one of the 7 stacks uses the same footprint and reaches 2 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 2 in every cell would mean checking 64 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.

Variant 7 easy answer: 5

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 3 by 2 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 3 by 2 block.
The front view is 2, 3 cells high, left to right.
The side view is 2, 1, 3 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

2 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 2

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

5

5 stacks are possible.

Answer: 5

4 · Reviewdoes it hold up?

Every one of the 5 stacks uses the same footprint and reaches 3 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 3 in every cell would mean checking 729 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.

Variant 8 medium answer: 3

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 3 by 2 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 3 by 2 block.
The front view is 2, 3 cells high, left to right.
The side view is 3, 1, 3 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

2 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 2

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

3

3 stacks are possible.

Answer: 3

4 · Reviewdoes it hold up?

Every one of the 3 stacks uses the same footprint and reaches 3 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 3 in every cell would mean checking 729 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.

Variant 9 hard answer: 5

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 2 by 3 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 2 by 3 block.
The front view is 4, 1, 4 cells high, left to right.
The side view is 3, 4 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

2 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 2

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

5

5 stacks are possible.

Answer: 5

4 · Reviewdoes it hold up?

Every one of the 5 stacks uses the same footprint and reaches 4 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 4 in every cell would mean checking 4096 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.

Variant 10 hard answer: 7

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 3 by 2 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 3 by 2 block.
The front view is 2, 4 cells high, left to right.
The side view is 3, 3, 4 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

0 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 0

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

7

7 stacks are possible.

Answer: 7

4 · Reviewdoes it hold up?

Every one of the 7 stacks uses the same footprint and reaches 4 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 4 in every cell would mean checking 4096 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.

Variant 11 hard answer: 7

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 3 by 2 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 3 by 2 block.
The front view is 4, 2 cells high, left to right.
The side view is 4, 4, 4 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

0 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 0

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

7

7 stacks are possible.

Answer: 7

4 · Reviewdoes it hold up?

Every one of the 7 stacks uses the same footprint and reaches 4 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 4 in every cell would mean checking 4096 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.

Variant 12 hard answer: 7

Unit cubes are to be stacked so that the views from above, the front and the side look like this. How many different stacks are possible?

Show solution

1 · Understandwhat's really being asked

A stack on a 3 by 3 footprint has to show the given front and side views. We want how many different stacks do.

Givens

The footprint is the 3 by 3 block.
The front view is 1, 4, 2 cells high, left to right.
The side view is 3, 4, 4 cells high, front to back.

Unknowns

How many different stacks match all three views.

Constraints

Every cell of the footprint holds at least one cube.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #17 Visualize Spatial Relationships#2 Make a Systematic List

Each view caps a whole line, so a cell is capped twice -- by its column and by its row. Where the smaller cap is 1 the height is forced; do all of those first and only then list what is still open.

3 · Execute4 carry out the plan

1Cap every cell twice

#17 Visualize Spatial Relationships 6.G.A.4

A cell can be no taller than its column's front view nor than its row's side view.

\text{cap} = \min(\text{front}, \text{side})

Two limits on every stack.

2Take the forced cells first

#3 Eliminate Possibilities 7.G.A.3

3 cells have a cap of 1, so their height is settled before anything is chosen.

\text{forced} = 3

Fewer cells left to decide.

3List what the rest can be

#2 Make a Systematic List 6.G.A.4

Each remaining cell can take any height up to its cap, but every column and row must actually reach its view -- which is what rules most combinations out.

\text{front and side must be attained}

Only a few survive.

4Count the survivors

#2 Make a Systematic List 6.G.A.4

Each surviving assignment is a different stack.

7

7 stacks are possible.

Answer: 7

4 · Reviewdoes it hold up?

Every one of the 7 stacks uses the same footprint and reaches 4 at its tallest -- the height both views agree on.

Another way: Trying every height from 1 to 4 in every cell would mean checking 262144 combinations; capping each cell first is what makes the list short.

Standardsmin grade 7

6.G.A.4 Surface area using nets of 3D figures — Reading each view as a cap on a whole line of stacks.
7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures — Deciding which cells the views leave open.

💡Takeaway. Do the parts that have only one answer first. What is left is a much shorter list.