SOLUTION: Different numbers can be made using the digits 1, 5, 6, 8 and a decimal point. How many possibilities are there if each digit must be used once and the decimal point must lie bet

There are 4! = 4∙3∙2∙1 = 24 ways to form a 4-digit whole number without a decimal.
Then there are 3 ways to place the decimal between two of the 4 digits.

That's 4!∙3 = 24∙3 = 72 different numbers.

Here are all 72:

1-3           1.568         15.68         156.8
4-6           1.586         15.86         158.6
7-9           1.658         16.58         165.8
10-12         1.685         16.85         168.5
13-15         1.856         18.56         185.6
16-18         1.865         18.65         186.5
19-21         5.168         51.68         516.8
22-24         5.186         51.86         518.6
25-27         5.618         56.18         561.8
28-30         5.681         56.81         568.1
31-33         5.816         58.16         581.6
34-36         5.861         58.61         586.1
37-39         6.158         61.58         615.8
40-42         6.185         61.85         618.5
43-45         6.518         65.18         651.8
46-48         6.581         65.81         658.1
49-51         6.815         68.15         681.5
52-54         6.851         68.51         685.1
55-57         8.156         81.56         815.6
58-60         8.165         81.65         816.5
61-63         8.516         85.16         851.6
64-66         8.561         85.61         856.1
67-69         8.615         86.15         861.5
70-72         8.651         86.51         865.1

Edwin