Question

How many different three-letter arrangements can be formed using the letters in the word ABSOLUTE if each letter is used only once?

Answer 1

Answer:

336 arrangements

Step-by-step explanation:

the word ABSOLUTE has 8 individual letters

we are asked to pick 3 letters out of 8 letters and we note that in this case, the order of the letters matter, hence we have to permutate 3 letters of 8

₈P₃ = 8! / (8-3)!

= 8! / 5!

= 336 arrangements

Edit: fixed spelling

Answer 2

We will find that there are 336 different 3-letter arrangements.

In the word ABSOLUTE we have a total of 8 different letters, from these, we want to select 3, and then we want to order the 3 letters.

So the number of different groups of 3 letters that we can take out of the set of 8 letters is:

$C(8, 3) = \frac{8!}{(8 - 3)!*3!} = \frac{8*7*6}{3*2} = 56$

Then we want to find the number of ways in which we can arrange these 3 letters.

Let's say that we have 3 places and 3 letters:

• For the first one, we have 3 options.
• For the second one, we have 2 options (one is already in the first place)
• For the third one, we have 1 option.

The number of permutations is given by the product between the numbers of options, we will get:

P = 3*2*1 = 6

The total number of different arrangements is equal to the product between the number of different sets of 3 letters that we can make, and the permutations of these 3 letters, so we have:

C = 6*56 = 336

If you want to learn more, you can read:

https://brainly.com/question/25606878