Problem: A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and $\textit{still}$ have at least one card of each color and at least one card with each number is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution: We recast this problem to coloring 8 cells in a 7x7 grid, such that each row and column has at least one colored cell. If a cell in row a and column b is colored, that means we drew a card with color a and number b.
By pigeonhole principle, there exists a row that has 2 colored cells, and intuitively, that row is unique. There also exists a column with 2 colored cells. Call the unique row, "Row A", and Call the unique column, "Column B".
If we are able to remove a card so that we still have a card of each color and number, that means, we can erase a cell, so each row still has at least one colored cell, and so does each column. If we erase a cell not in Row A, then the erased cell's row will be empty. If we erase a cell not in Column B, the erased cell's column will be empty. Thus the cell that we erase must be in both Row A and Column B. There is only one cell in both Row A and Column B, and thus that cell must be colored.
We first count the number of successful outcomes. We must chose which row to be "Row A", and which column to be "Column B". There are 7 ways to chose "Row A", and 7 ways to chose "Column B". Because of the logic used above, the intersection of Row A and Column B must be colored. However, Row A needs to have 2 colored cells. There are 6 other uncolored cells to chose from. Similarly, there are 6 uncolored cells in Column B.
Note that we can ignore all the rows and columns that contain colored cells. This is because Row A already contains 2 colored cells, and thus we can't color any more cells in Row A, similarly for Column B. The other row that contains a colored cell is not Row A, so it can only contain 1 colored cell, which it has, thus we can't color anymore cells in it. Similarly for the other column.
We can actually remove these rows/columns. This leaves a 5x5 grid, full of uncolored cells. Each row can have 1 cell, and so can each column. We count the # of colorings from the perspective of the rows. We look at the first row, it has 5 choices. We can look at the 2nd row, it has 4 choices, as it's selected cell can't occupy the same column. We continue down the line, and see there are $5! = 120$ ways to color.
Thus, the # of successful outcomes is $5! \cdot 6 \cdot 6 \cdot 7 \cdot 7$
We can then count the # of possible outcomes.
We can count the # of unsuccessful outcomes. We select "Row A" and "Column B" as usual, with $7 \cdot 7$ ways. We don't color the intersection, so we have to color 2 cells in each. There are 6 other cells that can be colored in each, so we have ${6 \choose 2} = 15$ ways to color each. Then, we can get rid of rows and columns using the method above and we are left with a 4x4 grid. There are $4! = 24$ ways to color.
Thus, the # of unsuccessful outcomes is $4! \cdot 15 \cdot 15 \cdot 7 \cdot 7$
After some calculation, we get $\frac{successful}{possible} = \frac{successful}{successful+unsuccessful} = \frac{5! \cdot 6^2 \cdot 7^2}{7^2 \cdot(5! \cdot 6^2 + 4! \cdot 15^2} = \frac{5! \cdot 6^2}{5! \cdot 6^2 +4! \cdot 15^2} = \frac{5 \cdot 6^2}{5 \cdot 6^2 + 15^2} = \frac{180}{180+225} = \frac{4}{9} => 4+9 = \boxed{13}$

Answer:  13