2. You might like to refer to Example 6.5.5 on page 327 of the textbook.
Solution 1: The following table illustrates some solutions to the equation x + y + z = 11.
Categories for x, y and z |
Solution to the equation x + y + z = 11 |
| ×× | ××××××××× | | x = 2, y = 9, z = 0 |
| ×××× | ×××× | ××× | x = 4, y = 4, z = 3 |
| × | ××× | ××××××× | x = 1, y = 3, z = 7 |
The table demonstrates that the number of solutions to the equation x + y + z = 11 is the same as the number of arrangements of 11 crosses and 2 bars.
From the previous section we know that if there are 13 objects, of which 11 are of one type and 2 are of another type, then the number of distinct arrangements of the 13 objects is:
| 13! | = |
13 · 12 · 11! | = |
78. |
| 11! 2! | 11! · 2 · 1 |
Solution 2: Think of a bag containing at least 11 objects of each of 3 different types (x, y, z). (The reason we need at least 11 of each type is so that we could choose all objects the same type.) Now choose 11 objects from the bag (these objects are the crosses in the table above). This is a problem involving r-combinations with repetition allowed, where n = 3 and r = 11.
| The number of solutions is | ( |
11 + 3 - 1 | ) | = |
( |
13 | ) | = |
13 · 12 · 11! | = |
78. |
| 11 | 11 | 11! · 2 · 1 |