#$&*
course Mth 152
9/9 3
Question:
`q001. Note that there are 14 questions in this assignment.
List all possible 3-letter 'words' that can be formed from the set of letters { a, b, c } without repeating any of the letters. Possible 'words' include 'acb' and 'bac'; however 'aba' is not permitted here because the letter 'a' is used twice (i.e., repeated).
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Abc, cba, bac, acb, cab, bca
3
I should list the possible solutions is a systematic way so in this case it would be alphabetical. When listing in this way your will be sure not to duplicate answers
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Question:
`q002. List all possible 3-letter 'words' that can be formed from the set of letters { a, b, c } if we allow repetition of letters. Possible 'words' include 'acb' and 'bac' as before; now 'aba' is permitted, as is 'ccc'.
Also specify how many words you listed, and how you could have figured out the result without listing all the possibilities.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Aaa, aab, aac, aba, abb, abc, acc, acb, aca, so there are 9 words that start with the letter A
Bbb, bba, bbc, baa, bab, bac, bca, bcb, bcc so there are 9 words that start with the letter B
Caa, Cab, Cac, Cba, Cbb, cbc, cca, ccb, ccc so there are 9 words that start with the letter C
We can then add 9+9+9= 27 so there are 27 possible words if we can repeat letters
3
List should be in systematic form
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Question:
`q003. If we form a 3-letter 'word' from the set {a, b, c}, not allowing repetitions, then:
How many choices do we have for the first letter chosen?
How many choices do we then have for the second letter?
How many choices do we therefore have for the 2-letter 'word' formed by the first two letters chosen?
How many choices are then left for the third letter?
How many choices does this make for the 3-letter 'word'?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
For the first letter chosen we will have 6 possibilities because each of the three possible choices there are only two that can follow so 3*2= 6
For the second letter we would have we would still have 6 possibilities
There is six possibilities for the 2- letter ‘word’ formed by the first two letters chosen
The third letter would only have one possibility left
3*2*1= 6 so there are 6 possible choices for each 3- letter ‘word’
3
Ok
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Question:
`q004. Check your answer to the last problem by listing the possibilities for the first two letters. Does your answer to that question match your list?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
first two letters could be ab, ac, ba, bc, ca or cb
after determining the first two letters the third is which everyone is left for example if your first two letters are cb the last must me a
the three letter word will then be abc, acb, bac, bca, cab and cba
3
Ok
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Question:
`q005. If we form a 3-letter 'word' from the set {a, b, c}, allowing repetitions, then
How many choices do we have for the first letter chosen?
How many choices do we then have for the second letter?
How many choices do we therefore have for the 2-letter 'word' formed by the first two letters chosen?
How many choices are then left for the third letter?
How many choices does this make for the 3-letter 'word'?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
We can have a total of three choices for the first letter chosen
We would still have three choices for the second letter
We would have 3 choices for the third letter as well
We can figure out that we have 27 possibilities because we would do 3*3*3=27
3
Ok
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Question:
`q006. If we were to form a 3-letter 'word' from the set {a, b, c, d}, without allowing a letter to be repeated, then
How many choices would we have for the first letter chosen?
How many choices would we then have for the second letter?
How many choices would we therefore have for the 2-letter 'word' formed by the first two letters chosen?
How many choices would then be left for the third letter?
How many possibilities does this make for the 3-letter 'word'?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
The first letter we would have 4 choices
We would have 3 for the second letter as well
We would have 2 choices for the third letter chosen
4*3*2= 24 possible choices
3
Ok
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Question:
`q007. List the 3-letter 'words' you can form from the set {a, b, c, d}, without allowing repetition of letters within a word. Does your list confirm your answer to the preceding question?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
abc, abd, acb, acdb, adb, adc
bac, bad, bca, bcd, bda, bdc;
cab, cad, cba, cbd, cda, cdb;
dab, dac, dba, dbc, dca, dcb.
4*6= 24 possibilities
3
Ok
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Question:
`q008. Imagine three boxes:
The first contains a set of billiard balls numbered 1 through 15.
The second contains a set of letter tiles with one tile for each letter of the alphabet.
The third box contains colored rings, one for each color of the rainbow (these colors are red, orange, yellow, green, blue, indigo and violet, abbreviated ROY G BIV).
If one object is chosen from each box, how many possibilities are there for the collection of objects chosen?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
15 * 26 * 7
3
Ok
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Question:
`q009. For the three boxes of the preceding problem, how many of the possible 3-object collections contain an odd number?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
8 * 26 * 7 = 1456
3
Ok
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Question:
`q010. For the three boxes of the preceding problem, how many of the possible collections contain an odd number and a vowel?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
8 * 5 * 7 = 280
3
Ok
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Question:
`q011. For the three boxes of the preceding problem, how many of the possible collections contain an even number, a consonant and one of the first three colors of the rainbow?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
7 * 21 * 3 = 441
3
Ok
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Question:
`q012.
For the three boxes of the preceding problem, how many of the possible collections contain an even number or a vowel?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
1274 + 525 - 245 = 1555
3
Ok
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Question:
`q013. For the three boxes of the preceding problems, if we choose two balls from the first box, then a tile from the second and a ring from the third, how many possible collections are there?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
15 * 14 * 26 * 7
3
Ok
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Question:
`q014. For the three boxes of the preceding problems, if we choose only from the first box, and choose three balls, how many possible ways are there to make our choice?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
5 * 14 * 13 / 6
3
Ok
"
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