. Consider a (quite rudimentary) language whose alphabet only contains four letters (say: A, B, C, and D). a) How many three-letter words may exist in this language? b) How many three-letter words can be formed if letter repetition is disallowed, i. e., no letter can appear more than once in any given word? c) Would your answer for (b) change in the case of four-letter words? Explain.

a. There are 24 three letter word in the language.

b. There are 24 three letter word in the language without repetition.

c. No. It wouldn't change because the answer would still be the same when calculated.

Step-by-step explanation:

a. How many three-letter words may exist in this language?

This is a permutation problem. And permutation is given as:

P(n, r) = n! / (n - r)!

So we solve 4 Permutation 3 i.e

P(4, 3) = 4! / (4 - 3)!

P(4, 3) = 4! / 1!

P(4, 3) = (4 * 3 * 2 * 1 ) / 1

P(4, 3) = 24

There are 24 three letter word in the language.

b. How many three-letter words can be formed if letter repetition is disallowed, i.e., no letter can appear more than once in any given word?

This is a combination problem. And combination is given as:

C(n, r) = n! / r!(n - r)!

Since repetition is not allowed, the possible three letter word in the language will be:

4C1 * 3C1 * 2C1

4C1 = 4! / 1!(4-1)!

4C1 = 4! / 1!3!

4C1 = 24/6 = 4

3C1 = 3!/1!(3-1)!

3C1 = 3!/1!2! = 6/2 = 3

2C1 = 2!/1!(2-1)! = 2!/1!1! = 2

The possible three letter word in the language will be:

4 * 3 * 2 = 24

c. Would your answer for (b) change in the case of four-letter words? Explain

It wouldn't change because the answer would still be the same when calculated. It will be:

4C1 * 3C1 * 2C1 * 1C1

= 4 * 3 * 2 * 1 = 24

The reason it doesn't change is that after calculating for three letters without repetition, we are left with only 1 letter.

first one because it can't have a fraction in it

step-by-step explanation: