Ways to place $5$ Ps in a $4 × 4$ grid so that each row has at least one P? 
In how many ways can $5$ P's be placed in $16$ identical squares formed by a $4$ × $4$ grid, such that each row has at least one P?

The number of ways of introducing first P will be: $16$ ( let us say the chosen cell is in first row)
Next P has $12$ cells to go in. (let us put that in $2$nd row)
$3$rd P has $8$ cells. (put that in $3$rd row)
$4$th P has $4$ cells. (put that in $4$th row)
And because $4$ cells out of $16$ have been full now, so the final P has $12$ cells to go in.
Making the total number of ways to place = $16 \cdot 12 \cdot 8 \cdot 4 \cdot 12 = 73728$ ways.
Where am I going wrong?
 A: Please note all $P$'s are identical so if there was no restriction, all you would do is to choose $5$ cells from $16$ cells, which is $ \displaystyle {16 \choose 5}$.
As there are $5$ P's and each row must have at least one $P$, one of the rows will have two $P$'s and rest three rows will have one $P$ each.
For each row with one $P$, there are $4$ columns to choose from and for the row with two $P$, there are $4 \choose 2 ~ $ ways to choose two columns. Also there are $4$ ways to choose the row with two $P$'s.
That leads to $ ~ \displaystyle 4^4 \cdot {4 \choose 2} ~$ ways.
A: Suppose we put the $P$'s in the same place you did, but I prefer to start in the fourth row and working up. The result will look the same, yet you did not take this into account in your computations. The order in which we place the $P$'s does not matter, only the cells in which we place them.
We can solve this by looking row per row (you added an example, but did not chose your example 'ad random'):

*

*How many ways to put a $P$ in the first row?

*How many ways to put a $P$ in the second row?

*How many ways to put a $P$ in the third row?

*How many ways to put a $P$ in the fourth row?

*How many ways to place the remaining $P$ in the remaining squares?

*How many ways to switch the two $P$'s that are together in a row? This amount should be divided out, since we double counted those possibilities (as order of placing the $P$'s does not matter).

