number of three-letter code words from the letters of the word WARDAYA

Question

Find the number of three-letter code words that can be formed from the letters of the word WARDAYA.

I am stuck on the part of the repetition of As. I know how to use permutations on the distinct letters, but I'm stuck after that. I do not how to find the final answer with the $3$ As.

I researched this kind of questions, and I found out I need to find the distinct letters first and ignore the repeated ones first. (In this case, there are $3$ As). WARDAYA has $5$ distinct letters, so $P(5,3) = 5 \times 4 \times 3 = 60$ ways (No repetition). I do not know what I should do after this.

@Audrey I suggest removing the "I don't have a clue on how to do this" comment. Explain that you know how to handle the case in which no letters are repeated but are stuck on the case in which letters are repeated. — N. F. Taussig, Jun 07 '18 at 09:43
I think coded words mean like AAA, ARD, DRA .etc @Vinyl_coat_jawa — Audrey, Jun 07 '18 at 09:45
is "wardayaWARDAYA" the actual word or is a typo? I mean, the point is that "w" and "W" are different? — Simone Masiero, Jun 07 '18 at 09:48
@SimoneMasiero That was an error introduced in the editing process. The "word" is WARDAYA. — N. F. Taussig, Jun 07 '18 at 09:51
What's $ used for here ? It's my first time posting a question here @N. F. Taussig — Audrey, Jun 07 '18 at 10:20
Dollar signs are used to place what is between them in math mode. This MathJax tutorial explains how to typeset mathematics on this site. — N. F. Taussig, Jun 07 '18 at 10:22
You can follow this approach: https://math.stackexchange.com/questions/20238/6-letter-permutations-in-mississippi — Muralidharan, Jun 07 '18 at 10:46

score 0 · Accepted Answer · answered Jun 07 '18 at 09:46

0

Strategy: A three-letter code from the letters of the word WARDAYA can have up to three As. Break into cases depending on the number of repeated letters.

No repeated letters: There are five different letters. You have to arrange three of them. As you found, there are $P(5, 3) = 60$ ways to do this.
The letter A is used twice: Choose the two positions occupied by the As. Choose which of the other four letters will be placed in the remaining position.
The letter A is used three times: AAA is the only code word of this type.

Since these three cases are mutually exclusive and exhaustive, the answer can be obtained by adding the results for the three cases.

answered Jun 07 '18 at 09:46

N. F. Taussig

76,571

Is the second strategy, P(4,3) x P(3,2) ? – Audrey Jun 07 '18 at 10:44
No. There are $\binom{3}{2}$ ways to choose the positions of the two As and four ways to choose one of the other letters to go in the remaining open position, so you should have $\binom{3}{2}\binom{4}{1} = 3 \cdot 4 = 12$ possible code words for the second case. They are AAD, AAR, AAW, AAY, ADA, ARA, AWA, AYA, DAA, RAA, WAA, YAA. When you see the word choose, you should be thinking in terms of combinations. – N. F. Taussig Jun 07 '18 at 10:58

score 0 · Answer 2 · answered Jun 07 '18 at 10:51

The problem can be solved in following way: There would be three classes/type of permutation

When every entry is "A", i.e. AAA, which implies there is only 1 such combination.
When there are 2 A's, which can be in three way: AAX, AXA , XAA ; where X is one of the four letters in the word( W,R,D,Y), all possible permutation are 4 * 3 = 12
When there are no identical letters , i.e. only 1 A is allowed, rest two excluded, which has 60 possible permutations.

Summing all: 1 + 12 + 60 = 73

number of three-letter code words from the letters of the word WARDAYA

2 Answers2