A five-digit positive integer is created using each of the odd digits $1,3,5,7,9$ once so that

- the thousands digit is larger than the hundreds digit,
- the thousands digit is larger than the ten thousands digit,
- the tens digit is larger than the hundreds digit, and
- the tens digit is larger than the units digit.

How many such five-digit positive integers are there?
(A) 12
(B) 8
(C) 16
(D) 14
(E) 10
We write such a five-digit positive integer with digits $V W X Y Z$.

We want to count the number of ways of assigning $1,3,5,7,9$ to the digits $V, W, X, Y, Z$ in such a way that the given properties are obeyed.

From the given conditions, $W>X, W>V, Y>X$, and $Y>Z$.

The digits 1 and 3 cannot be placed as $W$ or $Y$, since $W$ and $Y$ are larger than both of their neighbouring digits, while 1 is smaller than all of the other digits and 3 is only larger than one of the other possible digits.

The digit 9 cannot be placed as $V, X$ or $Z$ since it is the largest possible digit and so cannot be smaller than $W$ or $Y$. Thus, 9 is placed as $W$ or as $Y$.

Therefore, the digits $W$ and $Y$ are 9 and either 5 or 7 .

Suppose that $W=9$ and $Y=5$. The number is thus $V 9 X 5 Z$.

Neither $X$ or $Z$ can equal 7 since $7>5$, so $V=7 . X$ and $Z$ are then 1 and 3 or 3 and 1 .

There are 2 possible integers in this case.

Similarly, if $Y=9$ and $W=5$, there are 2 possible integers.

Suppose that $W=9$ and $Y=7$. The number is thus $V 9 X 7 Z$.

The digits 1, 3, 5 can be placed in any of the remaining spots. There are 3 choices for the digit $V$. For each of these choices, there are 2 choices for $X$ and then 1 choice for $Z$.

There are thus $3 \times 2 \times 1=6$ possible integers in this case.

Similarly, if $Y=9$ and $W=7$, there are 6 possible integers.

Overall, there are thus $2+2+6+6=16$ possible integers.