Problem #113 says:

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a “bouncy” number; for example, 155349.

As n increases, the proportion of bouncy numbers below n increases such that there are only 12951 numbers below one-million that are not bouncy and only 277032 non-bouncy numbers below 10^(10).

How many numbers below a googol (10^(100)) are not bouncy?

This is a simple combinatorial counting problem. As such we just need to calculate the binomial of a few numbers and remove the duplicates. The binomial requires the use of a unbounded integer class (in this case the biginteger class from System.Numerics) if you don’t want to use such a class you need to do some optimization on the binomial calculation.

Solution below in FSharp requires .Net 4 the FSharp powerpack. Runs in under 1 second.

#light

open System
open System.Diagnostics
open Microsoft.FSharp.Linq
open Microsoft.FSharp.Collections
open Microsoft.FSharp.Math
open System.Numerics

let rec Factorial (x:BigInteger) =
    if x = BigInteger.One then
        BigInteger.One
    else
        BigInteger.Multiply (x,(Factorial (BigInteger.op_Decrement x)))

let binomial (n:int) (k:int) =
    let bN = BigInteger n
    let bK = BigInteger k
    (Factorial bN) / ((Factorial bK) * (Factorial(bN-bK)))

let watch = new Stopwatch()
watch.Start()

let BigBinom = binomial 110 10
let SmallBinom = binomial 109 9
let n100 = BigInteger (10*100)
let two = BigInteger 2

Console.WriteLine(BigBinom + SmallBinom - n100 - two)
watch.Stop()
Console.WriteLine(watch.Elapsed)
Console.ReadKey() |> ignore