Number of permutations of four letter word using letters anything between 0-2 times each

Question

The first part of the problem involves calculating the number of four letter words possible to form from five letters A, B, C, D and E. Each letter can be used 0, 1 or 2 times. The second part of the question involves calculating the probability of randomly drawing such a four letter word having A:s as the first two letters.

I have googled on similar problems but do have difficulties understanding the logic, so if you have any recommendations on sites dealing with combinations/permutations of words (and in general) it would be greatly appreciated.

Answer 1

First part

The problem can be simplified by analysing the three possible cases: no double letters, a single double letter and two double letters.

• Case 1: No double letters

  This is a simple case, you only need to choose 4 letters from a pool of 5 letters where order matters:

$$\frac{5!}{1!}=5 \cdot 4 \cdot 3 \cdot 2=120$$

• Case 2: A single double letter

  This time you need to choose a double letter and two single letters. You can choose any of the 5 letters to be a double letter, one of the remaining 4 letters as a single letter and one of the 3 remaining letters as the last single letter. Remember that choosin letter $X$ first and $Y$ second or $Y$ first and $X$ second is equivalent for the single letters, so the number of combination of letters we can choose for the word are $\frac{5 \cdot 4 \cdot 3}{2}= 30$.

  Keep in mind that the order you pick the letters matters as long as they are not the same letter.

  $$\frac{4!}{1! \cdot 1! \cdot 2!}=12$$

  $30 \cdot 12=360$ cases.

• Case 3: Two double letters

  You can choose $\frac{5 \cdot 4}{2}= 10$ pairs of different letters. Now you only need to count how many permutations you can build with two pairs:

  $$\frac{4!}{2! \cdot 2!}=6$$

  $10 \cdot 6=60$ cases.

• Total combinations:

  $120+360+60=540$

Second part

For the second part, you need to distinguish two cases, the third and fourth letters are different or the are the same:

• Two different letters: Applying the same logic as above, you need to choose two different letters out of the remaining 4 letters (B, C, D and E) where the order matters:

  $$4 \cdot 3=12$$

• A double letter: This is a trivial case, you must only choose a letter out of 4 letters: 4 cases.

• Probability: The probability of obtaining a word that starts with two $A$s is equal to the number of words that start with two $A$s divided by the total ammount of cases:

  $$P=\frac{12+4}{540}=\frac{4}{135}$$