Mathematics, 23.08.2020 21:01 kell22wolf

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

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Answer from: camp591

Choice C. $3,\! 780$ .

Step-by-step explanation:

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

Four $\verb!E!$ 's, Two $\verb!N!$ 's, Two $\verb!S!$ 's, andOne $\verb!T!$ .

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:

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

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 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:

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

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 ( $4! \times 2! \times 2! \times 1! = 96$ ) to find the number of arrangements that are truly unique:

$\begin{aligned} & 362,\!880 \times \frac{1}{4! \times 2! \times 2! \times 1!} = 3,\!780\end{aligned}$ .

Answer from: Quest

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

$y\propto x$

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

Answer from: Quest

answer: it's d

step-by-step explanation:

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