code: def solve():
      """
      In England the currency is made up of pound, P, and pence, p, and there are eight coins in general circulation:
  
      1p, 2p, 5p, 10p, 20p, 50p, P1 (100p) and P2 (200p).
      It is possible to make P2 in the following way:
  
      1 P1 + 1 50p + 2 20p + 1 5p + 1 2p + 3 1p
      How many different ways can P2 be made using any number of coins?
      """
  
      coins = [1, 2, 5, 10, 20, 50, 100, 200]
  
      def balance(pattern):
          return sum(
              coins[x] * pattern[x] for x in range(0, len(pattern))
          )
  
      def gen(pattern, coinnum, num):
          coin = coins[coinnum]
          for p in range(0, num / coin + 1):
              newpat = pattern[:coinnum] + (p,)
              bal = balance(newpat)
              if bal > num:
                  return
              elif bal == num:
                  yield newpat
              elif coinnum < len(coins) - 1:
                  for pat in gen(newpat, coinnum + 1, num):
                      yield pat
  
      return sum(1 for pat in gen((), 0, 200))
  
label_desc: Use image node_id 1 for calling a global function with example usage: solve() and returns: sum, sum