, 2025

We begin by recognizing that in the given list, each of the
digits $1$ through $7$ occurs at least once as a units digit,
and at least once as a tens digit.

For example, the digit $1$ occurs
twice as a units digit ($11$ and
$31$), and three times as a tens
digit ($11$, $12$ and $14$).

Counting the number of times each of the digits $1$ through $7$ occurs as a units digit and as a tens
digit, we get:

Digit
$1$
$2$
$3$
$4$
$5$
$6$
$7$

Number of times occurring as a units digit
$2$
$1$
$1$
$4$
$1$
$2$
$1$

Number of times occurring as a tens digit
$3$
$1$
$1$
$3$
$1$
$2$
$1$

The units digit of each number in the list matches the tens digit of
the number that follows it.

This tells us that if we ignore the tens digit of the first number in
the list and the units digit of the last number in the list, then the
number of times that each digit occurs as a units digit must be equal to
the number of times that it occurs as a tens digit.

Looking back to the table above, we see that this is true for all digits
except 1 and 4.

Since the digit 1 occurs twice as a units digit and three times as a
tens digit, then the tens digit of the first number in the list must be
equal to 1.

Similarly, the digit 4 occurs four times as a units digit and three
times as a tens digit, and so the units digit of the last number in the
list must be equal to 4.

Ignoring the number 14 for a moment, we separate the 11 remaining
numbers into two distinct lists, which we call $A$ and $B$. $$A: 11,12,23,31 \ \ \ \ \ \ B: 44,45,46,56,64,67,74$$ Each digit in
$A$ is less than or equal to $3$, and each digit in $B$ is greater than or equal to $4$.

Since $14$ is the only number given
that does not appear in $A$ or $B$, and $14$ has a digit that appears in $A$ and a digit that appears in $B$, then $14$ is the only number that can ’connect’
the numbers in $A$ to those in $B$.

Further, this tells us that the numbers in $A$ must be arranged and then placed before
an arrangement of the numbers in $B$,
with $14$ appearing between the two
arrangements.

Also, the arrangement of the numbers in $A$ must begin and end with a $1$, and the arrangement of the numbers in
$B$ must begin and end with a $4$ (since $14$ occurs between the two lists).

Next, we count the number of different ways to arrange the numbers in
$A$, starting and ending with $1$.

We begin by recognizing that each of the digits $2$ and $3$ occurs exactly once as a units digit
and once as a tens digit, and so $12$, $23$, $31$ must appear together in this order
(the two $2$s must occur together and
the two $3$s must occur
together).

The list must begin and end with a $1$, and so there are $2$ possible locations for the $11$ and thus $2$ possible arrangements of the numbers in
$A$: $11,
12, 23, 31$, and$12, 23, 31,
11$.

Next, we count the number of different ways to arrange the numbers in
$B$, starting and ending with $4$.

We begin by recognizing that each of the digits $5$ and $7$ occurs exactly once as a units digit
and once as a tens digit, and so $45$, $56$ must appear together in this order
(the two $5$s must occur together),
and $67$, $74$ must appear together in this order
(the two $7$s must occur
together).

The arrangement ends with a $4$, and
thus cannot end with $45$, $56$, and so at least one more number must
immediately follow $45$, $56$.

There are two such possibilities: $45, 56,
64$, and$45, 56, 67, 74$
(recall that $67, 74$ must remain
together), which leads to exactly two distinct cases to consider.

Case 1: $\boldsymbol{45, 56, 64}$ occur
together in this order.

In this case, the remaining numbers are $44$, $46$, $67$, and $74$.

Since $44$ has two equal digits, its
location in the arrangement of the $B$ list cannot change the first digit in
the list (which must be $4$), and
cannot change the last digit in the list (which must also be $4$), and thus we ignore $44$ for the moment.

The remaining numbers, $46$, $67$, $74$ must occur together in this order. Can
you see why?

Since the blocks $45, 56, 64$ and
$46, 67, 74$ must each occur together
in their respective orders, this gives two possible arrangements of the
$B$ list (ignoring the $44$).

These are: $45, 56, 64, 46, 67, 74$,
and $46, 67, 74, 45, 56, 64$.

Next, we determine the number of different ways to place $44$ into each of these arrangements.

In the $45, 56, 64, 46, 67, 74$
arrangement, the $44$ may appear at
the start, at the end, or between the $64$ and $46$, which gives $3$ different arrangements of the $B$ list. (These are: $44, 45, 56, 64, 46, 67, 74$, and $45, 56, 64, 46, 67, 74, 44$, and $45, 56, 64, 44, 46, 67, 74$.)

In the $46, 67, 74, 45, 56, 64$
arrangement, the $44$ may appear at
the start, at the end, or between the $74$ and $45$, which gives $3$ more arrangements of the $B$ list, or $6$ in total for Case 1.

Case 2: $\boldsymbol{45, 56, 67, 74}$
occur together in this order.

In this case, the remaining numbers are $44$, $46$, and $64$.

We again begin by ignoring $44$ for
the moment.

The remaining numbers, $46, 64$ must
occur together in this order.

Since the blocks $45, 56, 67, 74$ and
$46, 64$ must each occur together in
their respective orders, this gives two possible arrangements of the
$B$ list (ignoring the $44$).

These are: $45, 56, 67, 74, 46, 64$
and $46, 64, 45, 56, 67, 74$.

In the $45, 56, 67, 74, 46, 64$
arrangement, the $44$ may appear at
the start, at the end, or between the $74$ and $46$, which gives $3$ more different arrangements of the
$B$ list.

In the $46, 64, 45, 56, 67, 74$
arrangement, the $44$ may appear at
the start, at the end, or between the $64$ and $45$, which gives $3$ more arrangements of the $B$ list, or $6$ in total for Case 2.

Thus, there are a total of $6+6=12$ different ways to arrange the
$B$ list.

There are $2$ different ways to
arrange the numbers in list $A$,
$12$ different ways to arrange the
numbers in list $B$, and exactly
$1$ way to place the number $14$ between arrangements of each of the
two lists. Thus, the total number of arrangements of the given list is
$2\times12\times1=24$.