How many ways can you make 2 pounds given these sizes of coins:

1p, 2p, 5p, 10p, 20p, 50p, 1£ and 2£

Where 1£ is 100p.

We can take either 0 or 1 2£ coin.

answer <- 0

for(a in seq(0,200,by=200)){
  for(b in seq(0,(200-a),by=100)){
    for(c in seq(0,(200-a-b), by=50)){
      for(d in seq(0,(200-a-b-c),by=20)){
        for(e in seq(0,(200-a-b-c-d), by=10)){
          for(f in seq(0,(200-a-b-c-d-e), by=5)){
            for(g in seq(0,(200-a-b-c-d-e-f),by=2)){
              answer <- answer + 1
            }
          }
        }
      }
    }
 }
}

How to explain it…

I need to count up to 200p total.

a is the part of the 200p that is made up by a 2£ coint. That is either 0 or 200. Or the sequence from 0 to 200 taken in increments of 200.

What is left after considering 2£ coins, is 200 minus the value of a. The number of 1£ coins we can use to make up that amount is all the values from 0 to what is left. Ie 0 to 200-a in increments of 100. The part of the total 200p that is made up by 1£ coins is b.

After adding some 1£ coins, there are 200-a-b left. That can be made by 50p coins. The part of what is left i c, and that is values from 0 to 200-a-b in increments of 50, ie seq(0,(200-a-b), by=50)

Etc.

The trick is to make sure that the ranges of the inner loops is controlled by what happened in previous loops.

Otherwise I will loop over far too many values.