Calculate The Combinations Of Entries

(I am unable to use Colo’s HTML Maker to post this—I have noticed several newer members indicating the same thing).

|....| 1 | 2 | 3 |4 | 5 | 6 | 7 | 8 |
| A | Y | N | Y | N | Y | Y | N | N |
| B | Y | N | N | Y | Y | N | N | N |
| C | Y | Y | N | Y | N | Y | Y | N |
| D | Y | Y | Y | N | Y | Y | N | Y |

The answer to the table above is: 144 combinations (using "Y" entries only). The answer is not to do just a simple count of the entries in each row and multiply as this produces 360 combinations rather than the desired 144 combinations. In this example, combinations such as 1-1-1-1 or 5-4-4-3 would not need to be calculated, which is the reason the answer is not 360 combinations.

The table always allows for 8 x 4 cell entries.
It is possible that every cell can be selected (the answer would be 8!/4! in this instance);
At least one entry must occur in each of the first 3 rows. In this example, if row 4 was excluded the answer would be 38 combinations (3 rows);

I am looking for 2 formulae: The first one to calculate the combinations of entries in 4 rows and the second to calculate for 3 rows only.