How Many Ways Can You Make A Pound With Old Imperial Coins?

Question

How Many Ways Can You Make A Pound With Old Imperial Coins?

2 Answers

Answer Question

Answer 1

It all depends which coins were in circulation at the time you want your answer to be correct for.

For starters and by way of comparison for an American dollar using pennies, nickels, dimes and quarters there are 242 ways. Add in the half-dollar and you get to 292. Add another one if you want to include the dollar coin.

For decimal sterling using 1p, 2p, 5p, 10p, 20p and 50p there are 4562 ways to make a pound. Again, add one if you want to include the pound coin itself.

With old imperial coins because a pound is 480 times a halfpenny the combinations really start to add up.

Just before decimalisation the denominations in circulation were halfpenny, penny, thrupenny bit, sixpence, shilling and florin. This gives 2,023,428 possibilities.

Go back to when the farthing was still around and the numbers become truly impressive rising to 156,844,190 permutations.

If you add in the half-crown and crown you get to 362,091,949

Consider also the half sovereign and sovereign and you can add another 5,858,634 further ways.

Of course you might also choose to include other coins like the double florin, groat and half-farthing. If you did all these the result would be many billions.

Crazy isn't it?

NB I can't guarantee I haven't made some logical or arithmetical error in these calculations so best not to treat them as gospel.

Maddog

Answer 2

In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?

There are more than 4000 possibilities so when you try this question you will find that counting all possibilities is too tedious unless you have a good system to reduce the work and a good notation to record work in progress. If you are going to get the answer you will need to find a good method which you can explain clearly.

Here is one method you might like to follow. Using a spreadsheet saves work but it is still easy to do without one.

Use the notation 100(1,2,5,10,20,50) for the number of combinations of the listed smaller coins which make up one pound and similarly for smaller amounts, for example 30(1,2,5) is the number of combinations of 1p, 2p and 5p coins which make up 30p.

Step 1 Show that the number of ways of changing X pence into 1p and 2p coins is (X/2 + 1) when X is even and (X + 1)/2 when X is odd. Now fill in column A in the table below.

Step 2 Fill in column B in the table using the results in column A and using the earlier results as you work your way down the column. For example we can make up 10 pence using no 5p coins, or one 5p coin or two 5p coins, hence:

10(1,2,5) = 10(1,2) + 5(,1,2) + 1 = 6 + 3 + 1 = 10

and similarly to make up 20 pence we use zero, one, two, three or four 5p coins giving:

20(1,2,5) = 20(1,2) + 15(1,2) + 10(1,2) + 5(1,2) + 1 = 11 + 8 + 6 + 3 + 1 = 29

Step 3 Fill in column C where, for example, corresponding to zero, one, two and three 10p coins we get:

30(1,2,5,10) = 30(1,2,5) + 20(1,2,5) + 10(1,2,5) + 1 = 58 + 29 + 10 + 1 = 98

Step 4 Now you should be able to continue in this manner to fill in the whole table and to get the answer in the bottom right hand corner.

Table showing the numbers of combinations of smaller coins to make up the amounts shown:
A B C D E
5(1,2)=
10(1,2)= 10(1,2,5)=10 10(1,2,5,10)= 10(1,2,5,10,20)= 10(1,2,5,10,20,50)=
15(1,2)=
20(1,2)= 20(1,2,5)=29 20(1,2,5,10)= 20(1,2,5,10,20)= 20(1,2,5,10,20,50)=
25(1,2)=
30(1,2)= 30(1,2,5)=58 30(1,2,5,10)=98 30(1,2,5,10,20)= 30(1,2,5,10,20,50)=
35(1,2)=
40(1,2)= 40(1,2,5)= 40(1,2,5,10)= 40(1,2,5,10,20)= 40(1,2,5,10,20,50)=
45(1,2)=
50(1,2)= 50(1,2,5)= 50(1,2,5,10)= 50(1,2,5,10,20)= 50(1,2,5,10,20,50)=
55(1,2)=
60(1,2)= 60(1,2,5)= 60(1,2,5,10)= 60(1,2,5,10,20)= 60(1,2,5,10,20,50)=
65(1,2)=
70(1,2)= 70(1,2,5)= 70(1,2,5,10)= 70(1,2,5,10,20)= 70(1,2,5,10,20,50)=
75(1,2)=
80(1,2)= 80(1,2,5)= 80(1,2,5,10)= 80(1,2,5,10,20)= 80(1,2,5,10,20,50)=
85(1,2)=
90(1,2)= 90(1,2,5)= 90(1,2,5,10)= 90(1,2,5,10,20)= 90(1,2,5,10,20,50)=
95(1,2)=
100(1,2)= 100(1,2,5)= 100(1,2,5,10)= 100(1,2,5,10,20)= 100(1,2,5,10,20,50)=

There are other ways to do this and you might like to find a different method of your own, perhaps writing a computer program to find the result, do let us know. It would be cool to publish several different methods.
So using imperial coins the answer will be very much greater.