Question

Lamia has the letter cards A, Z, D, Y, and E in a bag. If she selects a permutation of the cards at random, what is the probability that she will spell the word "ZAYED”?

Answer 1

`\displaystyle (1)/(5!) = (1)/(120) \approx 0.00833`.

Explanation:

Note that all five letters here are distinct (i.e., none of them is repeated.) There are
`\displaystyle P(5,\, 5) = (5!)/((5 - 5)!) = 5! = 120` ways to arrange five distinct items (where the order of the arrangement matters.)

The reason is that there are five choices for the first item, four choices for the second item, three choices for the third item, etc. Hence, the numerator is
`5 * 4* 3 * 2 * 1`, which is the same as
`5!`. On the other hand, since there's only one way to choose five items out of five (i.e., to select them all,) the denominator would be
`1`.

Note that the
`\verb!ZAYED!` is just one of that
`5!` possible permutations. If the cards are arranged in random, all these permutations ought to have an equal probability. Therefore:

`\begin{aligned}& P(\verb!ZAYED!) \\ &= \frac{\text{Number of permutations that gives $\texttt{ZAYED}$}}{\text{Number of all permutations involved}} \\ &= (1)/(5!) = (1)/(120) \approx 0.00833\end{aligned}`.