← 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

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Generated variants — 12

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

Variant 1 easy

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?

Your answer

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

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?

Your answer

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

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?

Your answer

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

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?

Your answer

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

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?

Your answer

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

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?

Your answer

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

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?

Your answer

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

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?

Your answer

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

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?

Your answer

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

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?

Your answer

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

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?

Your answer

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

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?

Your answer

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.

Where next?

← Back to the problem: Use the conditions that force a value first, then list what is left