Greater and Greater!

Relevant wiki: Combinations - Problem Solving

The 5 distinct numbers can range from 1 to 29. Let's first not worry about the order of these numbers, because we could sort them later. Now, between B and C there is one "free spot" that will be "lost" (cannot be occupied by any number), so there are only 28 spots to fill. Similarily we will lose 2 and 3 spots between C and D, and D and E respectively. So there are only 23 spots to fill with 5 numbers which can be expressed as ${23 \choose 5} = 33649$ .

Edit: Realized too late this approach is very similar to the one of @Brian Charlesworth. Sorry, no intention to take your credit.

Define the following functions:

• $A(n) =$ The number of ways you can pick $A$ less than $n+1$
• $B(n) =$ The number of ways you can pick $A$ and $B$ with $B < n+1$
• $C(n) =$ The number of ways you can pick $A$ , $B$ , and $C$ with $C < n+1$
• $D(n) =$ The number of ways you can pick $A$ , $B$ , $C$ , and $D$ with $D < n+1$
• $E(n) =$ The number of ways you can pick $A$ , $B$ , $C$ , $D$ , and $E$ with $E < n+1$

Then,

• $A(n) = n$
• $B(1) = 0$
• $B(n>1) = A(n-1) + B(n-1)$
• $C(n<4) = 0$
• $C(n>3) = B(n-2) + C(n-1)$
• $D(n<7) = 0$
• $D(n>6) = C(n-3) + D(n-1)$
• $E(n<11) = 0$
• $E(n>10) = D(n-4) + E(n-1)$

Solving for these you get:

$E(29) = \boxed{33649}$