Practice · Count functions by multiplying output choices · Sensim Math

Variant 1 easy

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ even for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4\}$ for each input $1$ , $2$ , $3$ . Count the choices where every input plus its output is even.

Givens

Inputs: $1$ , $2$ , $3$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ even for $a=1,2,3$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into three small ones: list the allowed outputs for $f(1)$ , $f(2)$ and $f(3)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn even sums into matching parity

#7 Identify Subproblems 8.F.A.1

A sum is even exactly when both numbers are odd or both are even. So

f(a)

must have the same parity as

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{1,3\}

:

2

choices. Input

2

is even, so

f(2)\in\{2,4\}

:

2

choices. Input

3

is odd, so

f(3)\in\{1,3\}

:

2

choices.

Among

1

to

4

there are two odd numbers and two even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

and

f(3)

, made independently.

2\times2\times2=8

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

8

4 · Reviewdoes it hold up?

Without the condition there are $4\times4\times4=64$ functions. The condition keeps roughly half of each input's choices, and $8$ is a small, sensible part of $64$ .

Another way: Fix $f(2)$ first. It has $2$ allowed values, and each one leaves $4$ ways to choose $f(1)$ , $f(3)$ . Total $2\times4=8$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Variant 2 easy

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ , $4$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ odd for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4\}$ for each input $1$ , $2$ , $3$ , $4$ . Count the choices where every input plus its output is odd.

Givens

Inputs: $1$ , $2$ , $3$ , $4$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ odd for $a=1,2,3,4$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into four small ones: list the allowed outputs for $f(1)$ , $f(2)$ , $f(3)$ and $f(4)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn odd sums into opposite parity

#7 Identify Subproblems 8.F.A.1

A sum is odd exactly when one number is odd and the other is even. So

f(a)

must have the opposite parity to

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{2,4\}

:

2

choices. Input

2

is even, so

f(2)\in\{1,3\}

:

2

choices. Input

3

is odd, so

f(3)\in\{2,4\}

:

2

choices. Input

4

is even, so

f(4)\in\{1,3\}

:

2

choices.

Among

1

to

4

there are two odd numbers and two even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

,

f(3)

and

f(4)

, made independently.

2\times2\times2\times2=16

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

16

4 · Reviewdoes it hold up?

Without the condition there are $4\times4\times4\times4=256$ functions. The condition keeps roughly half of each input's choices, and $16$ is a small, sensible part of $256$ .

Another way: Fix $f(2)$ first. It has $2$ allowed values, and each one leaves $8$ ways to choose $f(1)$ , $f(3)$ , $f(4)$ . Total $2\times8=16$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Variant 3 easy

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ , $5$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ odd for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4,5\}$ for each input $1$ , $2$ , $3$ . Count the choices where every input plus its output is odd.

Givens

Inputs: $1$ , $2$ , $3$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ , $5$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ odd for $a=1,2,3$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into three small ones: list the allowed outputs for $f(1)$ , $f(2)$ and $f(3)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn odd sums into opposite parity

#7 Identify Subproblems 8.F.A.1

A sum is odd exactly when one number is odd and the other is even. So

f(a)

must have the opposite parity to

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{2,4\}

:

2

choices. Input

2

is even, so

f(2)\in\{1,3,5\}

:

3

choices. Input

3

is odd, so

f(3)\in\{2,4\}

:

2

choices.

Among

1

to

5

there are three odd numbers and two even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

and

f(3)

, made independently.

2\times3\times2=12

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

12

4 · Reviewdoes it hold up?

Without the condition there are $5\times5\times5=125$ functions. The condition keeps roughly half of each input's choices, and $12$ is a small, sensible part of $125$ .

Another way: Fix $f(2)$ first. It has $3$ allowed values, and each one leaves $4$ ways to choose $f(1)$ , $f(3)$ . Total $3\times4=12$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Variant 4 easy

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ , $5$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ even for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4,5\}$ for each input $1$ , $2$ , $3$ . Count the choices where every input plus its output is even.

Givens

Inputs: $1$ , $2$ , $3$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ , $5$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ even for $a=1,2,3$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into three small ones: list the allowed outputs for $f(1)$ , $f(2)$ and $f(3)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn even sums into matching parity

#7 Identify Subproblems 8.F.A.1

A sum is even exactly when both numbers are odd or both are even. So

f(a)

must have the same parity as

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{1,3,5\}

:

3

choices. Input

2

is even, so

f(2)\in\{2,4\}

:

2

choices. Input

3

is odd, so

f(3)\in\{1,3,5\}

:

3

choices.

Among

1

to

5

there are three odd numbers and two even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

and

f(3)

, made independently.

3\times2\times3=18

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

18

4 · Reviewdoes it hold up?

Without the condition there are $5\times5\times5=125$ functions. The condition keeps roughly half of each input's choices, and $18$ is a small, sensible part of $125$ .

Another way: Fix $f(2)$ first. It has $2$ allowed values, and each one leaves $9$ ways to choose $f(1)$ , $f(3)$ . Total $2\times9=18$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Variant 5 medium

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ , $4$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ , $5$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ odd for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4,5\}$ for each input $1$ , $2$ , $3$ , $4$ . Count the choices where every input plus its output is odd.

Givens

Inputs: $1$ , $2$ , $3$ , $4$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ , $5$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ odd for $a=1,2,3,4$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into four small ones: list the allowed outputs for $f(1)$ , $f(2)$ , $f(3)$ and $f(4)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn odd sums into opposite parity

#7 Identify Subproblems 8.F.A.1

A sum is odd exactly when one number is odd and the other is even. So

f(a)

must have the opposite parity to

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{2,4\}

:

2

choices. Input

2

is even, so

f(2)\in\{1,3,5\}

:

3

choices. Input

3

is odd, so

f(3)\in\{2,4\}

:

2

choices. Input

4

is even, so

f(4)\in\{1,3,5\}

:

3

choices.

Among

1

to

5

there are three odd numbers and two even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

,

f(3)

and

f(4)

, made independently.

2\times3\times2\times3=36

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

36

4 · Reviewdoes it hold up?

Without the condition there are $5\times5\times5\times5=625$ functions. The condition keeps roughly half of each input's choices, and $36$ is a small, sensible part of $625$ .

Another way: Fix $f(2)$ first. It has $3$ allowed values, and each one leaves $12$ ways to choose $f(1)$ , $f(3)$ , $f(4)$ . Total $3\times12=36$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Variant 6 medium

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ , $4$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ , $5$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ even for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4,5\}$ for each input $1$ , $2$ , $3$ , $4$ . Count the choices where every input plus its output is even.

Givens

Inputs: $1$ , $2$ , $3$ , $4$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ , $5$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ even for $a=1,2,3,4$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into four small ones: list the allowed outputs for $f(1)$ , $f(2)$ , $f(3)$ and $f(4)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn even sums into matching parity

#7 Identify Subproblems 8.F.A.1

A sum is even exactly when both numbers are odd or both are even. So

f(a)

must have the same parity as

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{1,3,5\}

:

3

choices. Input

2

is even, so

f(2)\in\{2,4\}

:

2

choices. Input

3

is odd, so

f(3)\in\{1,3,5\}

:

3

choices. Input

4

is even, so

f(4)\in\{2,4\}

:

2

choices.

Among

1

to

5

there are three odd numbers and two even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

,

f(3)

and

f(4)

, made independently.

3\times2\times3\times2=36

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

36

4 · Reviewdoes it hold up?

Without the condition there are $5\times5\times5\times5=625$ functions. The condition keeps roughly half of each input's choices, and $36$ is a small, sensible part of $625$ .

Another way: Fix $f(2)$ first. It has $2$ allowed values, and each one leaves $18$ ways to choose $f(1)$ , $f(3)$ , $f(4)$ . Total $2\times18=36$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Variant 7 medium

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ , $5$ , $6$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ odd for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4,5,6\}$ for each input $1$ , $2$ , $3$ . Count the choices where every input plus its output is odd.

Givens

Inputs: $1$ , $2$ , $3$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ , $5$ , $6$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ odd for $a=1,2,3$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into three small ones: list the allowed outputs for $f(1)$ , $f(2)$ and $f(3)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn odd sums into opposite parity

#7 Identify Subproblems 8.F.A.1

A sum is odd exactly when one number is odd and the other is even. So

f(a)

must have the opposite parity to

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{2,4,6\}

:

3

choices. Input

2

is even, so

f(2)\in\{1,3,5\}

:

3

choices. Input

3

is odd, so

f(3)\in\{2,4,6\}

:

3

choices.

Among

1

to

6

there are three odd numbers and three even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

and

f(3)

, made independently.

3\times3\times3=27

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

27

4 · Reviewdoes it hold up?

Without the condition there are $6\times6\times6=216$ functions. The condition keeps roughly half of each input's choices, and $27$ is a small, sensible part of $216$ .

Another way: Fix $f(2)$ first. It has $3$ allowed values, and each one leaves $9$ ways to choose $f(1)$ , $f(3)$ . Total $3\times9=27$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Variant 8 medium

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ , $5$ , $6$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ even for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4,5,6\}$ for each input $1$ , $2$ , $3$ . Count the choices where every input plus its output is even.

Givens

Inputs: $1$ , $2$ , $3$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ , $5$ , $6$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ even for $a=1,2,3$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into three small ones: list the allowed outputs for $f(1)$ , $f(2)$ and $f(3)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn even sums into matching parity

#7 Identify Subproblems 8.F.A.1

A sum is even exactly when both numbers are odd or both are even. So

f(a)

must have the same parity as

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{1,3,5\}

:

3

choices. Input

2

is even, so

f(2)\in\{2,4,6\}

:

3

choices. Input

3

is odd, so

f(3)\in\{1,3,5\}

:

3

choices.

Among

1

to

6

there are three odd numbers and three even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

and

f(3)

, made independently.

3\times3\times3=27

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

27

4 · Reviewdoes it hold up?

Without the condition there are $6\times6\times6=216$ functions. The condition keeps roughly half of each input's choices, and $27$ is a small, sensible part of $216$ .

Another way: Fix $f(2)$ first. It has $3$ allowed values, and each one leaves $9$ ways to choose $f(1)$ , $f(3)$ . Total $3\times9=27$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Variant 9 hard

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ , $4$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ , $5$ , $6$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ even for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4,5,6\}$ for each input $1$ , $2$ , $3$ , $4$ . Count the choices where every input plus its output is even.

Givens

Inputs: $1$ , $2$ , $3$ , $4$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ , $5$ , $6$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ even for $a=1,2,3,4$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into four small ones: list the allowed outputs for $f(1)$ , $f(2)$ , $f(3)$ and $f(4)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn even sums into matching parity

#7 Identify Subproblems 8.F.A.1

A sum is even exactly when both numbers are odd or both are even. So

f(a)

must have the same parity as

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{1,3,5\}

:

3

choices. Input

2

is even, so

f(2)\in\{2,4,6\}

:

3

choices. Input

3

is odd, so

f(3)\in\{1,3,5\}

:

3

choices. Input

4

is even, so

f(4)\in\{2,4,6\}

:

3

choices.

Among

1

to

6

there are three odd numbers and three even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

,

f(3)

and

f(4)

, made independently.

3\times3\times3\times3=81

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

81

4 · Reviewdoes it hold up?

Without the condition there are $6\times6\times6\times6=1296$ functions. The condition keeps roughly half of each input's choices, and $81$ is a small, sensible part of $1296$ .

Another way: Fix $f(2)$ first. It has $3$ allowed values, and each one leaves $27$ ways to choose $f(1)$ , $f(3)$ , $f(4)$ . Total $3\times27=81$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Variant 10 hard

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ even for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4,5,6,7\}$ for each input $1$ , $2$ , $3$ . Count the choices where every input plus its output is even.

Givens

Inputs: $1$ , $2$ , $3$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ even for $a=1,2,3$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into three small ones: list the allowed outputs for $f(1)$ , $f(2)$ and $f(3)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn even sums into matching parity

#7 Identify Subproblems 8.F.A.1

A sum is even exactly when both numbers are odd or both are even. So

f(a)

must have the same parity as

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{1,3,5,7\}

:

4

choices. Input

2

is even, so

f(2)\in\{2,4,6\}

:

3

choices. Input

3

is odd, so

f(3)\in\{1,3,5,7\}

:

4

choices.

Among

1

to

7

there are four odd numbers and three even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

and

f(3)

, made independently.

4\times3\times4=48

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

48

4 · Reviewdoes it hold up?

Without the condition there are $7\times7\times7=343$ functions. The condition keeps roughly half of each input's choices, and $48$ is a small, sensible part of $343$ .

Another way: Fix $f(2)$ first. It has $3$ allowed values, and each one leaves $16$ ways to choose $f(1)$ , $f(3)$ . Total $3\times16=48$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Variant 11 hard

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ odd for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4,5,6,7\}$ for each input $1$ , $2$ , $3$ . Count the choices where every input plus its output is odd.

Givens

Inputs: $1$ , $2$ , $3$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ odd for $a=1,2,3$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into three small ones: list the allowed outputs for $f(1)$ , $f(2)$ and $f(3)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn odd sums into opposite parity

#7 Identify Subproblems 8.F.A.1

A sum is odd exactly when one number is odd and the other is even. So

f(a)

must have the opposite parity to

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{2,4,6\}

:

3

choices. Input

2

is even, so

f(2)\in\{1,3,5,7\}

:

4

choices. Input

3

is odd, so

f(3)\in\{2,4,6\}

:

3

choices.

Among

1

to

7

there are four odd numbers and three even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

and

f(3)

, made independently.

3\times4\times3=36

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

36

4 · Reviewdoes it hold up?

Without the condition there are $7\times7\times7=343$ functions. The condition keeps roughly half of each input's choices, and $36$ is a small, sensible part of $343$ .

Another way: Fix $f(2)$ first. It has $4$ allowed values, and each one leaves $9$ ways to choose $f(1)$ , $f(3)$ . Total $4\times9=36$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Variant 12 hard

For two variables $x$ and $y$ , each value $1$ , $2$ , $3$ of $x$ must be matched to exactly one of the values $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ of $y$ ; call such a function $y=f(x)$ . How many of these functions make $a+f(a)$ even for every value $a$ of $x$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A function here is a choice of one output from $\{1,2,3,4,5,6,7,8\}$ for each input $1$ , $2$ , $3$ . Count the choices where every input plus its output is even.

Givens

Inputs: $1$ , $2$ , $3$ .
Possible outputs: $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ .
Each input gets exactly one output.

Unknowns

The number of functions with $a+f(a)$ even for $a=1,2,3$ .

Constraints

Outputs may repeat; two inputs can share an output.

2 · Planchoose the strategy

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

The condition for each input involves only that input's own output, so split the problem into three small ones: list the allowed outputs for $f(1)$ , $f(2)$ and $f(3)$ separately, then multiply because the choices are independent.

3 · Execute3 carry out the plan

1Turn even sums into matching parity

#7 Identify Subproblems 8.F.A.1

A sum is even exactly when both numbers are odd or both are even. So

f(a)

must have the same parity as

a

.

Odd plus odd and even plus even are even; odd plus even is odd.

2List the allowed outputs

#2 Make a Systematic List 8.F.A.1

Input

1

is odd, so

f(1)\in\{1,3,5,7\}

:

4

choices. Input

2

is even, so

f(2)\in\{2,4,6,8\}

:

4

choices. Input

3

is odd, so

f(3)\in\{1,3,5,7\}

:

4

choices.

Among

1

to

8

there are four odd numbers and four even numbers.

3Multiply the choices

#7 Identify Subproblems 7.SP.C.8

Each function is one choice for each of

f(1)

,

f(2)

and

f(3)

, made independently.

4\times4\times4=64

Like choosing an outfit: every shirt goes with every pair of pants, so the counts multiply.

Answer:

64

4 · Reviewdoes it hold up?

Without the condition there are $8\times8\times8=512$ functions. The condition keeps roughly half of each input's choices, and $64$ is a small, sensible part of $512$ .

Another way: Fix $f(2)$ first. It has $4$ allowed values, and each one leaves $16$ ways to choose $f(1)$ , $f(3)$ . Total $4\times16=64$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Treating a function as one output chosen for each input.
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation — Counting outcomes of independent choices with the multiplication rule.

💡Takeaway. If each input's rule only looks at its own output, count each input's choices and multiply.

Count functions by multiplying output choices · 12 practice problems

Generated variants — 12

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