What is the number of different ways the letters of the word 'vector' can be placed in $8$ boxes of the figure below, such that no row is empty.
My attempt:
I selected firstly $2$ letters out of $6$ and placed on row no. 2 and 3 and then used $4$ box in 1st row as $$\binom{6}{2} \cdot 2! \cdot \binom{4}{4} \cdot 4! + \binom{6}{3} \cdot 3! \cdot \binom{4}{3} \cdot 3! + \binom{6}{4} \cdot 4! \cdot \binom{4}{2} \cdot 2!$$
But the answer is $26 \cdot 6!$
On
Let we have three rows $(1),(2)$ and $(3)$
We have to arrange letter $\mathcal{VECTOR}$ in these boxes so that no box remain empty.
So we have arrange all $6$ different letters in $(1),(2)$ row or in $(1),(3)$ row.
So we have $\displaystyle \binom{8}{6}-2=26$ ways of selection.
So total arrangements is $\displaystyle 26\times 6!$
There are $\binom{8}{6}$ ways to choose the six cells in which the letters of the word VECTOR are placed and $6!$ ways to arrange them in those cells. However, these choices include cases in which the second or third row is empty (it is not possible for the first row to be empty since there are only two empty cells). We must subtract these cases from the total. There are two ways to choose an empty row and $6!$ ways to arrange the letters in the remaining six cells. Hence, the number of ways the letters of the word VECTOR may be placed in the cells so that no row is empty is $$\binom{8}{6}6! - \binom{2}{1}6!$$
You attempted to place some of the letters in the bottom two rows, then place the remaining letters in the top row. However, you did not choose where the letters would be placed in the bottom two rows.
Four letters in the bottom two rows and two letters in the top row: There are $\binom{6}{4}$ ways to select which four letters will be placed in the bottom two rows and $4!$ ways to arrange the selected letters in those rows. There are $\binom{4}{2}$ ways to choose which cells of the top row will be occupied by the remaining two letters and $2!$ ways to arrange those letters in the selected cells. Hence, there are $$\binom{6}{4}4!\binom{4}{2}2!$$ such arrangements, as you found.
Three letters in the bottom two rows and three letters in the top row: There must be two letters in one of the bottom two rows and one letter in the other. There are $\binom{2}{1}$ ways to choose which of the bottom two rows will receive two letters. There are $\binom{6}{2}$ ways to select the letters that will appear in this row and $2!$ ways to arrange the selected letters in those cells. There are $\binom{2}{1}$ ways to choose which cell will be filled in the row that receives one letter and $\binom{4}{1}$ ways to choose which letter will fill that cell. There are $\binom{4}{3}$ ways to choose which three of the four cells in the top row will be occupied by the remaining three letters, and $3!$ ways to arrange them in these cells. Hence, there are $$\binom{2}{1}\binom{6}{2}2!\binom{4}{1}\binom{2}{1}\binom{4}{3}3!$$ such arrangements.
Two letters in the bottom two rows and four letters in the top row: We must place one letter in each of the bottom two rows so that no row will be empty. There are six ways to choose which letter will appear in the bottom row and two ways to choose which cell of the bottom row it will occupy. There are five ways to choose which of the remaining letters will appear in the middle row and two ways to choose which cell of the middle row it will occupy. The remaining four letters can be arranged in the four cells of the top row in $4!$ ways. Hence, there are $$\binom{6}{1}\binom{2}{1}\binom{5}{1}\binom{2}{1}4!$$ such arrangements.
To obtain the total number of arrangements, add the results from these three cases.
Simpler approach: That said, if you want to try counting directly, it would be easier to choose the six occupied cells, then multiply by the $6!$ ways the letters can be arranged in those cells.
Four objects in the top row, one each in each of the remaining rows: There are $\binom{4}{4}\binom{2}{1}\binom{2}{1}$ ways to choose the occupied cells.
Three objects in the top row, two in the middle row, and one in the bottom row: There are $\binom{4}{3}\binom{2}{2}\binom{2}{1}$ ways to choose the occupied cells.
Three objects in the top row, one in the middle row, and two in the bottom row: There are $\binom{4}{3}\binom{2}{1}\binom{2}{2}$ ways to choose the occupied cells.
Two objects in the top row, two in each of the bottom two rows: There are $\binom{4}{2}\binom{2}{2}\binom{2}{2}$ ways to choose the occupied cells.
In total, we obtain $$\left[\binom{4}{4}\binom{2}{1}\binom{2}{1} + 2\binom{4}{3}\binom{2}{2}\binom{2}{1} + \binom{4}{2}\binom{2}{2}\binom{2}{2}\right]6!$$ possible arrangements.