SOLUTION: contest consists of finding all of the code words that can be formed from the letters in the name "ATARI." Assume that the letter A can be used twice, but the others at most once.
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Question 1203679: contest consists of finding all of the code words that can be formed from the letters in the name "ATARI." Assume that the letter A can be used twice, but the others at most once.
A) How many five-letter words can be formed?
B) How many two-letter words can be formed?
C) How many words can be formed?
Found 3 solutions by math_tutor2020, ikleyn, greenestamps:
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
Part (a)
If we could tell the "A"s apart, then there would be 5! = 5*4*3*2*1 = 120 different five-letter permutations.
But because we cannot tell the "A"s apart, we must divide by 2 which corrects the erroneous double-count.
We'll have 120/2 = 60 different five-letter words possible.
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Part (b)
We have these four possible cases:- The letter "A" is used twice
- The letter "A" is used exactly once in the 1st slot.
- The letter "A" is used exactly once in the 2nd slot.
- The letter "A" is not chosen at all.
Case 1 has exactly one outcome. That outcome is AA.
Case 2 has 3 choices for the 2nd slot, which means we have the 3 permutations AT, AR, AI
Case 3 also has 3 outcomes. The logic is similar to case 2.
Case 4 has 3*2 = 6 permutations of the letters {T,R,I}
Add up the results
1+3+3+6 = 13
There are 13 different two letter sequences
The entire list is- AA
- AT
- AR
- AI
- TA
- TR
- TI
- RA
- RT
- RI
- IA
- IT
- IR
A combinatorics calculator I recommend is this
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html
It computes the nPr value, and also lists the different permutations.
Refer to the "List Them" section on that page.
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Part (c)
The previous part calculated 13 different two-letter words.
Let's calculate how many three-letter sequences are possible.
We have these cases- The letter "A" is used twice.
- The letter "A" is used exactly once.
- The letter "A" is not chosen at all.
Case 1
There are 3 slots to pick from where the letters T, R, or I will go.
After that slot is chosen, the other two slots are locked in as "A"
Therefore, we have 3*3 = 9 ways to do case 1.
Case 2
There are 3 slots to pick for the letter "A".
We have two more slots and 3 more letters to pick from to fill said slots. We'll have 3*2 = 6 permutations of those other letters.
Overall we have 3*6 = 18 ways to carry out case 2.
Case 3
There are 3*2*1 = 6 permutations of the letters {T,R,I}
Add up the results
9+18+6 = 33
We find there are 33 different three-letter words.
I'll let you do the scratch work to determine how many four-letter words are possible.
You should find there are 60 different four-letter words possible.
To summarize we have...- 60 ways to form a five-letter word.
- 60 ways to form a four-letter word.
- 33 ways to form a three-letter word.
- 13 ways to form a two-letter word.
There are 60+60+33+13 = 166 different words possible when we can only select letters from the list {A,T,A,R,I} where "A" can be used at most twice, but the other letters at most once.
-------------------------------
Answers:
(a) 60
(b) 13
(c) 166
Answer by ikleyn(52787) (Show Source): You can put this solution on YOUR website!
.
I think that the answer 166 to question (c) in the post by @math_tutor2020 should be changed:
4 other 1-letter words should be added to it
166 + 4 = 170.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
1-letter words....
There are 4 different letters to choose from.
The number of 1-letter words is 4.
2-letter words....
If both letters A are used, there is only 1 word -- AA
If the letter A is not repeated, then there are 4 choices for the first letter and then 3 choices for the second, so there are 4*3=12 2-letter words without both letters A.
The number of 2-letter words is 1+12 =13.
3-letter words....
If both letters A are used, then the combination of letters can be AAI, AAR, or AAT. There are (3!)/(2!) = 3 arrangements of each of those, for a total of 9.
If both letters A are not used, then there are 4 choices for the first letter, 3 for the second, and 2 for the third, for a total of 4*3*2 = 24.
The number of 3-letter words is 9+24 = 33.
4-letter words....
If both letters A are used, then the combination of letters can be AAIR, AAIT, or AART. There are (4!)/(2!) = 12 arrangements of each of those, for a total of 36.
If both letters A are not used, then the letters are ATRI in any order; there are 4! = 24 of those.
The number of 4-letter words is 36+24 = 60.
5-letter words....
The letters are ATARI in any order, the number of arrangements is (5!)/(2!) = 60.
There are 60 5-letter words.
Summary:
1-letter words: 4
2-letter words: 13
3-letter words: 33
4-letter words: 60
5-letter words: 60
Total number of words: 4+13+33+60+60 = 170
ANSWERS:
A) 5-letter words: 60
B) 2-letter words: 13
C) total number of words: 170
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