Question

In how many different, distinguishable orders can the letters of the word "TENNESSEE" be arranged? (A)24 (B)362,880 (C)3,780

Answer 1

Choice C.
`3,\! 780`.

Explanation:

The word "
`\verb!TENNESSEE!`" contains nine letters:

• Four
  `\verb!E!`'s,
• Two
  `\verb!N!`'s,
• Two
  `\verb!S!`'s, and
• One
  `\verb!T!`.

If all these nine letters are unique, the number of possible arrangements would be:

`P(9, 9) = 9! = 362,\! 880`.

However, because some of these letters were not unique, a large number of that
`362,\!880` arrangements would be duplicates. What would be the exact number of duplicates? Start by considering giving each of the duplicate letters an index number. For example, consider the duplicates due to the letter "
`\verb!E!`". There are four
`\verb!E!`'s in
`\verb!TENNESSEE!`. Label them as
`\verb!E!_1`,
`\verb!E!_2`,
`\verb!E!_3`, and
`\verb!E!_4`:

`\begin{array}{ccccccccc}\verb!T! & \verb!E! & \verb!N! & \verb!N! & \verb!E! & \verb!S! & \verb!S! & \verb!E! & \verb!E! \\ & \uparrow & & & \uparrow & & & \uparrow & \uparrow \\ & \verb!E!_1 & & & \verb!E!_2 & & & \verb!E!_3 & \verb!E!_4\end{array}`.

These four
`\verb!E!` can be shuffled and rearranged to give a large number of duplicates:

`\begin{array}ccccccccc& \verb!T! & \verb!E! & \verb!N! & \verb!N! & \verb!E! & \verb!S! & \verb!S! & \verb!E! & \verb!E! \\ &&\uparrow&&&\uparrow&&&\uparrow &\uparrow\\ 1 & & \verb!E!_1 & & & \verb!E!_2 & & & \verb!E!_3 & \verb!E!_4\\ 2 & & \verb!E!_1 & & & \verb!E!_2 & & & \verb!E!_4 & \verb!E!_3 \\ 3 & & \verb!E!_1 & & & \verb!E!_4 & & & \verb!E!_3 & \verb!E!_2 \\\ \vdots & & \vdots & & & \vdots & & & \vdots&\vdots \\4! = 24&&\verb!E!_4 & & & \verb!E!_3 & & & \verb!E!_2 & \verb!E!_1 \end{array}`.

Each of these duplicate corresponds to a unique way for arranging four unique item in a row. There are
`4! = 24` ways to arrange four unique items in a row. Therefore, for each arrangement that is truly unique, the letter "
`\verb!E!`" alone would have caused the arrangement to be counted
`4! = 24` times if the nine letters were assumed to be unique.

At the same time, letters "
`\verb!N!`" and "
`\verb!S!`" would further exaggerate the count by a factor of
`2` each. On the other hand, the letter
`\verb!T!` appeared only once and would not create duplicates.

Overall, if the nine letters were assumed to be unique, each arrangement that is truly unique would have been counted:

`4! * 2! * 2! * 1! = 96\; \text{times}`.

The count based on the incorrect assumption (that the nine letters are all distinct) is
`362,\!880`. Divide that count by the factor of exaggeration (
`4! * 2! * 2! * 1! = 96`) to find the number of arrangements that are truly unique:

`\begin{aligned} & 362,\!880 * (1)/(4! * 2! * 2! * 1!) = 3,\!780\end{aligned}`.