Question

How many different $10$-letter words (real or imaginary) can be formed from the following letters? $\text{O}$, $\text{Z}$, $\text{Q}$, $\text{Y}$, $\text{N}$, $\text{U}$, $\text{Z}$, $\text{J}$, $\text{W}$, $\text{V}$ ten-letter words (real or imaginary) can be formed with the given letters. (Type a whole number.)

Answer 1

Find the total number of letters formed form $\mathbf{10}$ words.

The total $10$ given letters are $\text{O}$, $\text{Z}$, $\text{Q}$, $\text{Y}$, $\text{N}$, $\text{U}$, $\text{Z}$, $\text{J}$, $\text{W}$, $\text{V}$.

The $\mathit{\text{n}}$ elements are rearranging then the number of rearrangements would be $\text{n}!$.

If some element $\text{a}$, $\text{b}$, and $\text{c}$ etc. repeats itself then the rearrangements would be $\frac{\mathit{\text{n}}!}{\mathit{\text{a}}!\times \mathit{\text{b}}!\times \mathit{\text{c}}!...}$.

If the total number of elements in the given problem is $10$ and $\mathit{\text{z}}$ repeated $2$ times, then the arrangements would be:

$\begin{array}{rcl}\frac{10!}{2!}& =& \frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2!}{2!}\\ & \Rightarrow & \frac{10!}{2!}=\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3}{1}\\ & \Rightarrow & \frac{10!}{2!}=10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\\ & \Rightarrow & \frac{10!}{2!}=1814400\end{array}$

Hence, the number of different $10$-letter words (real or imaginary) that can be formed is $\mathbf{1814400}$.