← Count numbers built from digit cards · Build the Largest or Smallest Value from Digit Cards

Count numbers built from digit cards · 11 practice problems

4.NBT.A.2

Generated variants — 11

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

Variant 1 easy answer: 6

Using each number card exactly once, you want to make four-digit numbers that are greater than 50005000. Find how many such numbers can be made in all.

7 0 4 2\boxed{7}\ \boxed{0}\ \boxed{4}\ \boxed{2}

Show solution
1 · Understandwhat's really being asked

Using the four digit cards 7, 0, 4, 2 each exactly once, build four-digit numbers and count how many of them are greater than 5000.

Givens
  • The four digit cards are 7, 0, 4, 2.
  • Each card must be used exactly once to form a four-digit number.
  • We only count numbers that are greater than 5000.
Unknowns
  • How many four-digit numbers greater than 5000 can be made.
Constraints
  • Every number is four digits, so the thousands place cannot be 0.
  • The number must be strictly greater than 5000.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Whether a number beats the threshold is decided almost entirely by the leading digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.

3 · Execute3 carry out the plan

1Decide which card can lead the number

#7 Identify Subproblems 4.NBT.A.2
A four-digit number's size is decided first by its thousands (leading) digit, and 0 cannot lead a four-digit number. Comparing each possible leading card against 5000, the cards that can produce a number greater than 5000 are 7.
leading card⇒7\text{leading card} \Rightarrow 7
Comparing multi-digit numbers starts from the highest place, so the leading digit tells you most of the story.

2Count arrangements led by 7

#2 Make a Systematic List 4.NBT.A.2
Put 7 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 7024, which is still greater than 5000, so all 6 count.
3×2×1=6,7024>50003 \times 2 \times 1 = 6,\quad 7024 > 5000
Three different cards into three slots give 6 different orderings.

3Add the cases

#2 Make a Systematic List 4.NBT.A.2
Combine the separate, non-overlapping leading-digit cases.
6=66 = 6
Each leading digit gives its own list, so we simply add the counts.
Answer: 6
4 · Reviewdoes it hold up?

There are 18 four-digit numbers in all using these cards. Counting only those greater than 5000 gives 6, a believable fraction of the total.

Another way: Write out the full systematic list of all 18 four-digit numbers and cross off each one that is 5000 or less, then count what remains.

Standardsmin grade 4
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing each four-digit number against 5000 by reading the thousands place first.
💡Takeaway. Look at the biggest place first: only some cards can lead a number past the threshold, then just count the orderings!
Variant 2 easy answer: 6

Using each number card exactly once, you want to make four-digit numbers that are greater than 60006000. Find how many such numbers can be made in all.

5 2 8 0\boxed{5}\ \boxed{2}\ \boxed{8}\ \boxed{0}

Show solution
1 · Understandwhat's really being asked

Using the four digit cards 5, 2, 8, 0 each exactly once, build four-digit numbers and count how many of them are greater than 6000.

Givens
  • The four digit cards are 5, 2, 8, 0.
  • Each card must be used exactly once to form a four-digit number.
  • We only count numbers that are greater than 6000.
Unknowns
  • How many four-digit numbers greater than 6000 can be made.
Constraints
  • Every number is four digits, so the thousands place cannot be 0.
  • The number must be strictly greater than 6000.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Whether a number beats the threshold is decided almost entirely by the leading digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.

3 · Execute3 carry out the plan

1Decide which card can lead the number

#7 Identify Subproblems 4.NBT.A.2
A four-digit number's size is decided first by its thousands (leading) digit, and 0 cannot lead a four-digit number. Comparing each possible leading card against 6000, the cards that can produce a number greater than 6000 are 8.
leading card⇒8\text{leading card} \Rightarrow 8
Comparing multi-digit numbers starts from the highest place, so the leading digit tells you most of the story.

2Count arrangements led by 8

#2 Make a Systematic List 4.NBT.A.2
Put 8 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 8025, which is still greater than 6000, so all 6 count.
3×2×1=6,8025>60003 \times 2 \times 1 = 6,\quad 8025 > 6000
Three different cards into three slots give 6 different orderings.

3Add the cases

#2 Make a Systematic List 4.NBT.A.2
Combine the separate, non-overlapping leading-digit cases.
6=66 = 6
Each leading digit gives its own list, so we simply add the counts.
Answer: 6
4 · Reviewdoes it hold up?

There are 18 four-digit numbers in all using these cards. Counting only those greater than 6000 gives 6, a believable fraction of the total.

Another way: Write out the full systematic list of all 18 four-digit numbers and cross off each one that is 6000 or less, then count what remains.

Standardsmin grade 4
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing each four-digit number against 6000 by reading the thousands place first.
💡Takeaway. Look at the biggest place first: only some cards can lead a number past the threshold, then just count the orderings!
Variant 3 easy answer: 6

Using each number card exactly once, you want to make four-digit numbers that are greater than 70007000. Find how many such numbers can be made in all.

6 0 2 9\boxed{6}\ \boxed{0}\ \boxed{2}\ \boxed{9}

Show solution
1 · Understandwhat's really being asked

Using the four digit cards 6, 0, 2, 9 each exactly once, build four-digit numbers and count how many of them are greater than 7000.

Givens
  • The four digit cards are 6, 0, 2, 9.
  • Each card must be used exactly once to form a four-digit number.
  • We only count numbers that are greater than 7000.
Unknowns
  • How many four-digit numbers greater than 7000 can be made.
Constraints
  • Every number is four digits, so the thousands place cannot be 0.
  • The number must be strictly greater than 7000.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Whether a number beats the threshold is decided almost entirely by the leading digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.

3 · Execute3 carry out the plan

1Decide which card can lead the number

#7 Identify Subproblems 4.NBT.A.2
A four-digit number's size is decided first by its thousands (leading) digit, and 0 cannot lead a four-digit number. Comparing each possible leading card against 7000, the cards that can produce a number greater than 7000 are 9.
leading card⇒9\text{leading card} \Rightarrow 9
Comparing multi-digit numbers starts from the highest place, so the leading digit tells you most of the story.

2Count arrangements led by 9

#2 Make a Systematic List 4.NBT.A.2
Put 9 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 9026, which is still greater than 7000, so all 6 count.
3×2×1=6,9026>70003 \times 2 \times 1 = 6,\quad 9026 > 7000
Three different cards into three slots give 6 different orderings.

3Add the cases

#2 Make a Systematic List 4.NBT.A.2
Combine the separate, non-overlapping leading-digit cases.
6=66 = 6
Each leading digit gives its own list, so we simply add the counts.
Answer: 6
4 · Reviewdoes it hold up?

There are 18 four-digit numbers in all using these cards. Counting only those greater than 7000 gives 6, a believable fraction of the total.

Another way: Write out the full systematic list of all 18 four-digit numbers and cross off each one that is 7000 or less, then count what remains.

Standardsmin grade 4
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing each four-digit number against 7000 by reading the thousands place first.
💡Takeaway. Look at the biggest place first: only some cards can lead a number past the threshold, then just count the orderings!
Variant 4 medium answer: 6

Using each number card exactly once, you want to make four-digit numbers that are greater than 80008000. Find how many such numbers can be made in all.

9 0 4 7\boxed{9}\ \boxed{0}\ \boxed{4}\ \boxed{7}

Show solution
1 · Understandwhat's really being asked

Using the four digit cards 9, 0, 4, 7 each exactly once, build four-digit numbers and count how many of them are greater than 8000.

Givens
  • The four digit cards are 9, 0, 4, 7.
  • Each card must be used exactly once to form a four-digit number.
  • We only count numbers that are greater than 8000.
Unknowns
  • How many four-digit numbers greater than 8000 can be made.
Constraints
  • Every number is four digits, so the thousands place cannot be 0.
  • The number must be strictly greater than 8000.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Whether a number beats the threshold is decided almost entirely by the leading digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.

3 · Execute3 carry out the plan

1Decide which card can lead the number

#7 Identify Subproblems 4.NBT.A.2
A four-digit number's size is decided first by its thousands (leading) digit, and 0 cannot lead a four-digit number. Comparing each possible leading card against 8000, the cards that can produce a number greater than 8000 are 9.
leading card⇒9\text{leading card} \Rightarrow 9
Comparing multi-digit numbers starts from the highest place, so the leading digit tells you most of the story.

2Count arrangements led by 9

#2 Make a Systematic List 4.NBT.A.2
Put 9 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 9047, which is still greater than 8000, so all 6 count.
3×2×1=6,9047>80003 \times 2 \times 1 = 6,\quad 9047 > 8000
Three different cards into three slots give 6 different orderings.

3Add the cases

#2 Make a Systematic List 4.NBT.A.2
Combine the separate, non-overlapping leading-digit cases.
6=66 = 6
Each leading digit gives its own list, so we simply add the counts.
Answer: 6
4 · Reviewdoes it hold up?

There are 18 four-digit numbers in all using these cards. Counting only those greater than 8000 gives 6, a believable fraction of the total.

Another way: Write out the full systematic list of all 18 four-digit numbers and cross off each one that is 8000 or less, then count what remains.

Standardsmin grade 4
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing each four-digit number against 8000 by reading the thousands place first.
💡Takeaway. Look at the biggest place first: only some cards can lead a number past the threshold, then just count the orderings!
Variant 5 medium answer: 12

Using each number card exactly once, you want to make four-digit numbers that are greater than 30003000. Find how many such numbers can be made in all.

4 7 0 1\boxed{4}\ \boxed{7}\ \boxed{0}\ \boxed{1}

Show solution
1 · Understandwhat's really being asked

Using the four digit cards 4, 7, 0, 1 each exactly once, build four-digit numbers and count how many of them are greater than 3000.

Givens
  • The four digit cards are 4, 7, 0, 1.
  • Each card must be used exactly once to form a four-digit number.
  • We only count numbers that are greater than 3000.
Unknowns
  • How many four-digit numbers greater than 3000 can be made.
Constraints
  • Every number is four digits, so the thousands place cannot be 0.
  • The number must be strictly greater than 3000.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Whether a number beats the threshold is decided almost entirely by the leading digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.

3 · Execute4 carry out the plan

1Decide which card can lead the number

#7 Identify Subproblems 4.NBT.A.2
A four-digit number's size is decided first by its thousands (leading) digit, and 0 cannot lead a four-digit number. Comparing each possible leading card against 3000, the cards that can produce a number greater than 3000 are 4, 7.
leading card⇒4,7\text{leading card} \Rightarrow 4, 7
Comparing multi-digit numbers starts from the highest place, so the leading digit tells you most of the story.

2Count arrangements led by 4

#2 Make a Systematic List 4.NBT.A.2
Put 4 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 4017, which is still greater than 3000, so all 6 count.
3×2×1=6,4017>30003 \times 2 \times 1 = 6,\quad 4017 > 3000
Three different cards into three slots give 6 different orderings.

3Count arrangements led by 7

#2 Make a Systematic List 4.NBT.A.2
Put 7 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 7014, which is still greater than 3000, so all 6 count.
3×2×1=6,7014>30003 \times 2 \times 1 = 6,\quad 7014 > 3000
Three different cards into three slots give 6 different orderings.

4Add the cases

#2 Make a Systematic List 4.NBT.A.2
Combine the separate, non-overlapping leading-digit cases.
6+6=126 + 6 = 12
Each leading digit gives its own list, so we simply add the counts.
Answer: 12
4 · Reviewdoes it hold up?

There are 18 four-digit numbers in all using these cards. Counting only those greater than 3000 gives 12, a believable fraction of the total.

Another way: Write out the full systematic list of all 18 four-digit numbers and cross off each one that is 3000 or less, then count what remains.

Standardsmin grade 4
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing each four-digit number against 3000 by reading the thousands place first.
💡Takeaway. Look at the biggest place first: only some cards can lead a number past the threshold, then just count the orderings!
Variant 6 medium answer: 12

Using each number card exactly once, you want to make four-digit numbers that are greater than 40004000. Find how many such numbers can be made in all.

2 7 5 0\boxed{2}\ \boxed{7}\ \boxed{5}\ \boxed{0}

Show solution
1 · Understandwhat's really being asked

Using the four digit cards 2, 7, 5, 0 each exactly once, build four-digit numbers and count how many of them are greater than 4000.

Givens
  • The four digit cards are 2, 7, 5, 0.
  • Each card must be used exactly once to form a four-digit number.
  • We only count numbers that are greater than 4000.
Unknowns
  • How many four-digit numbers greater than 4000 can be made.
Constraints
  • Every number is four digits, so the thousands place cannot be 0.
  • The number must be strictly greater than 4000.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Whether a number beats the threshold is decided almost entirely by the leading digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.

3 · Execute4 carry out the plan

1Decide which card can lead the number

#7 Identify Subproblems 4.NBT.A.2
A four-digit number's size is decided first by its thousands (leading) digit, and 0 cannot lead a four-digit number. Comparing each possible leading card against 4000, the cards that can produce a number greater than 4000 are 5, 7.
leading card⇒5,7\text{leading card} \Rightarrow 5, 7
Comparing multi-digit numbers starts from the highest place, so the leading digit tells you most of the story.

2Count arrangements led by 5

#2 Make a Systematic List 4.NBT.A.2
Put 5 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 5027, which is still greater than 4000, so all 6 count.
3×2×1=6,5027>40003 \times 2 \times 1 = 6,\quad 5027 > 4000
Three different cards into three slots give 6 different orderings.

3Count arrangements led by 7

#2 Make a Systematic List 4.NBT.A.2
Put 7 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 7025, which is still greater than 4000, so all 6 count.
3×2×1=6,7025>40003 \times 2 \times 1 = 6,\quad 7025 > 4000
Three different cards into three slots give 6 different orderings.

4Add the cases

#2 Make a Systematic List 4.NBT.A.2
Combine the separate, non-overlapping leading-digit cases.
6+6=126 + 6 = 12
Each leading digit gives its own list, so we simply add the counts.
Answer: 12
4 · Reviewdoes it hold up?

There are 18 four-digit numbers in all using these cards. Counting only those greater than 4000 gives 12, a believable fraction of the total.

Another way: Write out the full systematic list of all 18 four-digit numbers and cross off each one that is 4000 or less, then count what remains.

Standardsmin grade 4
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing each four-digit number against 4000 by reading the thousands place first.
💡Takeaway. Look at the biggest place first: only some cards can lead a number past the threshold, then just count the orderings!
Variant 7 medium answer: 12

Using each number card exactly once, you want to make four-digit numbers that are greater than 40004000. Find how many such numbers can be made in all.

3 8 1 5\boxed{3}\ \boxed{8}\ \boxed{1}\ \boxed{5}

Show solution
1 · Understandwhat's really being asked

Using the four digit cards 3, 8, 1, 5 each exactly once, build four-digit numbers and count how many of them are greater than 4000.

Givens
  • The four digit cards are 3, 8, 1, 5.
  • Each card must be used exactly once to form a four-digit number.
  • We only count numbers that are greater than 4000.
Unknowns
  • How many four-digit numbers greater than 4000 can be made.
Constraints
  • Every number is four digits, so the thousands place cannot be 0.
  • The number must be strictly greater than 4000.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Whether a number beats the threshold is decided almost entirely by the leading digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.

3 · Execute4 carry out the plan

1Decide which card can lead the number

#7 Identify Subproblems 4.NBT.A.2
A four-digit number's size is decided first by its thousands (leading) digit, and 0 cannot lead a four-digit number. Comparing each possible leading card against 4000, the cards that can produce a number greater than 4000 are 5, 8.
leading card⇒5,8\text{leading card} \Rightarrow 5, 8
Comparing multi-digit numbers starts from the highest place, so the leading digit tells you most of the story.

2Count arrangements led by 5

#2 Make a Systematic List 4.NBT.A.2
Put 5 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 5138, which is still greater than 4000, so all 6 count.
3×2×1=6,5138>40003 \times 2 \times 1 = 6,\quad 5138 > 4000
Three different cards into three slots give 6 different orderings.

3Count arrangements led by 8

#2 Make a Systematic List 4.NBT.A.2
Put 8 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 8135, which is still greater than 4000, so all 6 count.
3×2×1=6,8135>40003 \times 2 \times 1 = 6,\quad 8135 > 4000
Three different cards into three slots give 6 different orderings.

4Add the cases

#2 Make a Systematic List 4.NBT.A.2
Combine the separate, non-overlapping leading-digit cases.
6+6=126 + 6 = 12
Each leading digit gives its own list, so we simply add the counts.
Answer: 12
4 · Reviewdoes it hold up?

There are 24 four-digit numbers in all using these cards. Counting only those greater than 4000 gives 12, a believable fraction of the total.

Another way: Write out the full systematic list of all 24 four-digit numbers and cross off each one that is 4000 or less, then count what remains.

Standardsmin grade 4
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing each four-digit number against 4000 by reading the thousands place first.
💡Takeaway. Look at the biggest place first: only some cards can lead a number past the threshold, then just count the orderings!
Variant 8 medium answer: 12

Using each number card exactly once, you want to make four-digit numbers that are greater than 50005000. Find how many such numbers can be made in all.

8 0 5 3\boxed{8}\ \boxed{0}\ \boxed{5}\ \boxed{3}

Show solution
1 · Understandwhat's really being asked

Using the four digit cards 8, 0, 5, 3 each exactly once, build four-digit numbers and count how many of them are greater than 5000.

Givens
  • The four digit cards are 8, 0, 5, 3.
  • Each card must be used exactly once to form a four-digit number.
  • We only count numbers that are greater than 5000.
Unknowns
  • How many four-digit numbers greater than 5000 can be made.
Constraints
  • Every number is four digits, so the thousands place cannot be 0.
  • The number must be strictly greater than 5000.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Whether a number beats the threshold is decided almost entirely by the leading digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.

3 · Execute4 carry out the plan

1Decide which card can lead the number

#7 Identify Subproblems 4.NBT.A.2
A four-digit number's size is decided first by its thousands (leading) digit, and 0 cannot lead a four-digit number. Comparing each possible leading card against 5000, the cards that can produce a number greater than 5000 are 5, 8.
leading card⇒5,8\text{leading card} \Rightarrow 5, 8
Comparing multi-digit numbers starts from the highest place, so the leading digit tells you most of the story.

2Count arrangements led by 5

#2 Make a Systematic List 4.NBT.A.2
Put 5 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 5038, which is still greater than 5000, so all 6 count.
3×2×1=6,5038>50003 \times 2 \times 1 = 6,\quad 5038 > 5000
Three different cards into three slots give 6 different orderings.

3Count arrangements led by 8

#2 Make a Systematic List 4.NBT.A.2
Put 8 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 8035, which is still greater than 5000, so all 6 count.
3×2×1=6,8035>50003 \times 2 \times 1 = 6,\quad 8035 > 5000
Three different cards into three slots give 6 different orderings.

4Add the cases

#2 Make a Systematic List 4.NBT.A.2
Combine the separate, non-overlapping leading-digit cases.
6+6=126 + 6 = 12
Each leading digit gives its own list, so we simply add the counts.
Answer: 12
4 · Reviewdoes it hold up?

There are 18 four-digit numbers in all using these cards. Counting only those greater than 5000 gives 12, a believable fraction of the total.

Another way: Write out the full systematic list of all 18 four-digit numbers and cross off each one that is 5000 or less, then count what remains.

Standardsmin grade 4
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing each four-digit number against 5000 by reading the thousands place first.
💡Takeaway. Look at the biggest place first: only some cards can lead a number past the threshold, then just count the orderings!
Variant 9 hard answer: 12

Using each number card exactly once, you want to make four-digit numbers that are greater than 50005000. Find how many such numbers can be made in all.

1 9 4 6\boxed{1}\ \boxed{9}\ \boxed{4}\ \boxed{6}

Show solution
1 · Understandwhat's really being asked

Using the four digit cards 1, 9, 4, 6 each exactly once, build four-digit numbers and count how many of them are greater than 5000.

Givens
  • The four digit cards are 1, 9, 4, 6.
  • Each card must be used exactly once to form a four-digit number.
  • We only count numbers that are greater than 5000.
Unknowns
  • How many four-digit numbers greater than 5000 can be made.
Constraints
  • Every number is four digits, so the thousands place cannot be 0.
  • The number must be strictly greater than 5000.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Whether a number beats the threshold is decided almost entirely by the leading digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.

3 · Execute4 carry out the plan

1Decide which card can lead the number

#7 Identify Subproblems 4.NBT.A.2
A four-digit number's size is decided first by its thousands (leading) digit, and 0 cannot lead a four-digit number. Comparing each possible leading card against 5000, the cards that can produce a number greater than 5000 are 6, 9.
leading card⇒6,9\text{leading card} \Rightarrow 6, 9
Comparing multi-digit numbers starts from the highest place, so the leading digit tells you most of the story.

2Count arrangements led by 6

#2 Make a Systematic List 4.NBT.A.2
Put 6 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 6149, which is still greater than 5000, so all 6 count.
3×2×1=6,6149>50003 \times 2 \times 1 = 6,\quad 6149 > 5000
Three different cards into three slots give 6 different orderings.

3Count arrangements led by 9

#2 Make a Systematic List 4.NBT.A.2
Put 9 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 9146, which is still greater than 5000, so all 6 count.
3×2×1=6,9146>50003 \times 2 \times 1 = 6,\quad 9146 > 5000
Three different cards into three slots give 6 different orderings.

4Add the cases

#2 Make a Systematic List 4.NBT.A.2
Combine the separate, non-overlapping leading-digit cases.
6+6=126 + 6 = 12
Each leading digit gives its own list, so we simply add the counts.
Answer: 12
4 · Reviewdoes it hold up?

There are 24 four-digit numbers in all using these cards. Counting only those greater than 5000 gives 12, a believable fraction of the total.

Another way: Write out the full systematic list of all 24 four-digit numbers and cross off each one that is 5000 or less, then count what remains.

Standardsmin grade 4
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing each four-digit number against 5000 by reading the thousands place first.
💡Takeaway. Look at the biggest place first: only some cards can lead a number past the threshold, then just count the orderings!
Variant 10 hard answer: 12

Using each number card exactly once, you want to make four-digit numbers that are greater than 60006000. Find how many such numbers can be made in all.

9 1 6 0\boxed{9}\ \boxed{1}\ \boxed{6}\ \boxed{0}

Show solution
1 · Understandwhat's really being asked

Using the four digit cards 9, 1, 6, 0 each exactly once, build four-digit numbers and count how many of them are greater than 6000.

Givens
  • The four digit cards are 9, 1, 6, 0.
  • Each card must be used exactly once to form a four-digit number.
  • We only count numbers that are greater than 6000.
Unknowns
  • How many four-digit numbers greater than 6000 can be made.
Constraints
  • Every number is four digits, so the thousands place cannot be 0.
  • The number must be strictly greater than 6000.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Whether a number beats the threshold is decided almost entirely by the leading digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.

3 · Execute4 carry out the plan

1Decide which card can lead the number

#7 Identify Subproblems 4.NBT.A.2
A four-digit number's size is decided first by its thousands (leading) digit, and 0 cannot lead a four-digit number. Comparing each possible leading card against 6000, the cards that can produce a number greater than 6000 are 6, 9.
leading card⇒6,9\text{leading card} \Rightarrow 6, 9
Comparing multi-digit numbers starts from the highest place, so the leading digit tells you most of the story.

2Count arrangements led by 6

#2 Make a Systematic List 4.NBT.A.2
Put 6 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 6019, which is still greater than 6000, so all 6 count.
3×2×1=6,6019>60003 \times 2 \times 1 = 6,\quad 6019 > 6000
Three different cards into three slots give 6 different orderings.

3Count arrangements led by 9

#2 Make a Systematic List 4.NBT.A.2
Put 9 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 9016, which is still greater than 6000, so all 6 count.
3×2×1=6,9016>60003 \times 2 \times 1 = 6,\quad 9016 > 6000
Three different cards into three slots give 6 different orderings.

4Add the cases

#2 Make a Systematic List 4.NBT.A.2
Combine the separate, non-overlapping leading-digit cases.
6+6=126 + 6 = 12
Each leading digit gives its own list, so we simply add the counts.
Answer: 12
4 · Reviewdoes it hold up?

There are 18 four-digit numbers in all using these cards. Counting only those greater than 6000 gives 12, a believable fraction of the total.

Another way: Write out the full systematic list of all 18 four-digit numbers and cross off each one that is 6000 or less, then count what remains.

Standardsmin grade 4
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing each four-digit number against 6000 by reading the thousands place first.
💡Takeaway. Look at the biggest place first: only some cards can lead a number past the threshold, then just count the orderings!
Variant 11 hard answer: 12

Using each number card exactly once, you want to make four-digit numbers that are greater than 60006000. Find how many such numbers can be made in all.

8 3 1 6\boxed{8}\ \boxed{3}\ \boxed{1}\ \boxed{6}

Show solution
1 · Understandwhat's really being asked

Using the four digit cards 8, 3, 1, 6 each exactly once, build four-digit numbers and count how many of them are greater than 6000.

Givens
  • The four digit cards are 8, 3, 1, 6.
  • Each card must be used exactly once to form a four-digit number.
  • We only count numbers that are greater than 6000.
Unknowns
  • How many four-digit numbers greater than 6000 can be made.
Constraints
  • Every number is four digits, so the thousands place cannot be 0.
  • The number must be strictly greater than 6000.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Whether a number beats the threshold is decided almost entirely by the leading digit, so I split the count by which card sits in the thousands place (a subproblem), then systematically list the arrangements of the remaining three cards.

3 · Execute4 carry out the plan

1Decide which card can lead the number

#7 Identify Subproblems 4.NBT.A.2
A four-digit number's size is decided first by its thousands (leading) digit, and 0 cannot lead a four-digit number. Comparing each possible leading card against 6000, the cards that can produce a number greater than 6000 are 6, 8.
leading card⇒6,8\text{leading card} \Rightarrow 6, 8
Comparing multi-digit numbers starts from the highest place, so the leading digit tells you most of the story.

2Count arrangements led by 6

#2 Make a Systematic List 4.NBT.A.2
Put 6 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 6138, which is still greater than 6000, so all 6 count.
3×2×1=6,6138>60003 \times 2 \times 1 = 6,\quad 6138 > 6000
Three different cards into three slots give 6 different orderings.

3Count arrangements led by 8

#2 Make a Systematic List 4.NBT.A.2
Put 8 first. The remaining three cards fill the other places in any order, giving 3 x 2 x 1 = 6 orderings. The smallest of them is 8136, which is still greater than 6000, so all 6 count.
3×2×1=6,8136>60003 \times 2 \times 1 = 6,\quad 8136 > 6000
Three different cards into three slots give 6 different orderings.

4Add the cases

#2 Make a Systematic List 4.NBT.A.2
Combine the separate, non-overlapping leading-digit cases.
6+6=126 + 6 = 12
Each leading digit gives its own list, so we simply add the counts.
Answer: 12
4 · Reviewdoes it hold up?

There are 24 four-digit numbers in all using these cards. Counting only those greater than 6000 gives 12, a believable fraction of the total.

Another way: Write out the full systematic list of all 24 four-digit numbers and cross off each one that is 6000 or less, then count what remains.

Standardsmin grade 4
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing each four-digit number against 6000 by reading the thousands place first.
💡Takeaway. Look at the biggest place first: only some cards can lead a number past the threshold, then just count the orderings!