4 -letter "words" are formed using the letters A, B, C, D, E, F, G. How many such words are possible for each of the following conditions?

(a) No condition is imposed.
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(b) No letter can be repeated in a word.
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(c) Each word must begin with the letter A.
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(d) The letter C must be at the end.
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(e) The second letter must be a vowel.
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a) no condition: 7x7x7x7 = 7^4 = 2401


b) no repeats: 7x6x5x4 = 840

c) begin with A, can I repeat the others ?
I will assume I can:
1x7x7x7 = 343

d) same as c)
7x7x7x1 = 343

e) 2nd letter is a vowel
7x2x7x7 = 686

In c), d) and e) I assumed that letters can be repeated, you did not say otherwise.
If they cannot be repeated , it is easy to change the calculations

Thank you! I was able to do A & B but c,d,& e I would figure out. thanks again

For some particular website, the password must start with the letter X. Then there must be 5 more characters, and each of those characters may be a numeric digit or a letter of the alphabet. How many different passwords are possible?

(a) No condition is imposed.

Well, since no condition is imposed, we can use any of the available letters in each position. So, for each position, there are 7 letters to choose from. Therefore, there would be 7 choices for the first letter, 7 choices for the second letter, 7 choices for the third letter, and 7 choices for the fourth letter. Multiplying these choices together, we get 7 * 7 * 7 * 7 = 2401 possible words.

(b) No letter can be repeated in a word.
Now things get a bit trickier. Since no letter can be repeated, for the first position, we still have 7 choices. But for the second position, we have only 6 choices left, since one letter has already been used. Similarly, for the third position, we have 5 choices left, and for the fourth position, we have 4 choices left. So, the total number of words would be 7 * 6 * 5 * 4 = 840 possible words.

(c) Each word must begin with the letter A.
Considering that each word must start with A, we have only 1 choice for the first position. For the second position, we have 6 choices left since A has already been used. For the third position, we have 5 choices left, and for the fourth position, we have 4 choices left. So, the total number of words would be 1 * 6 * 5 * 4 = 120 possible words.

(d) The letter C must be at the end.
In this case, we have 1 choice for the last position, which must be C. For the first position, we have 6 choices left since C cannot be used. For the second position, we have 6 choices left (including C), and for the third position, we have 5 choices left. So, the total number of words would be 6 * 6 * 5 * 1 = 180 possible words.

(e) The second letter must be a vowel.
Considering that the second letter must be a vowel (A or E), we have 2 choices for the second position. For the first position, we have 6 choices left since one vowel has already been used. For the third position, we have 5 choices left, and for the fourth position, we have 4 choices left. So, the total number of words would be 6 * 2 * 5 * 4 = 240 possible words.

(a) No condition is imposed: To find the possible number of 4-letter words without any condition, we need to find the total number of combinations possible. Since we have 7 different letters (A, B, C, D, E, F, G), we can choose any letter as the first, second, third, and fourth letter in a word, with repetition allowed. Therefore, the total number of words is:

Number of choices for the first letter: 7
Number of choices for the second letter: 7
Number of choices for the third letter: 7
Number of choices for the fourth letter: 7

Total number of words = 7 * 7 * 7 * 7 = 2401

Therefore, 2401 different words are possible without any conditions imposed.

(b) No letter can be repeated: To find the possible number of 4-letter words without any repeated letters, we need to eliminate the choices for each subsequent letter after choosing the previous letter.

Number of choices for the first letter: 7
Number of choices for the second letter (excluding the first chosen letter): 6
Number of choices for the third letter (excluding the first two chosen letters): 5
Number of choices for the fourth letter (excluding the first three chosen letters): 4

Total number of words = 7 * 6 * 5 * 4 = 840

Therefore, 840 different words are possible with no letter repetition.

(c) Each word must begin with the letter A: To find the possible number of 4-letter words starting with A, we have only one choice for the first letter. The remaining three letters can be chosen from the remaining 6 letters (B, C, D, E, F, G).

Number of choices for the first letter (A): 1
Number of choices for the second letter: 6
Number of choices for the third letter: 6
Number of choices for the fourth letter: 6

Total number of words = 1 * 6 * 6 * 6 = 216

Therefore, 216 different words are possible where each word starts with the letter A.

(d) The letter C must be at the end: Similar to the previous condition, we have only one choice for the last letter (C) and the remaining three letters can be chosen from the remaining 6 letters (A, B, D, E, F, G).

Number of choices for the first letter: 6
Number of choices for the second letter: 6
Number of choices for the third letter: 6
Number of choices for the fourth letter (C): 1

Total number of words = 6 * 6 * 6 * 1 = 216

Therefore, 216 different words are possible where the letter C is at the end.

(e) The second letter must be a vowel: To find the possible number of 4-letter words where the second letter is a vowel (A or E), we have 2 choices for the second letter and the remaining three letters can be chosen from the remaining 6 letters (B, C, D, F, G).

Number of choices for the first letter: 7
Number of choices for the second letter (A or E): 2
Number of choices for the third letter: 6
Number of choices for the fourth letter: 6

Total number of words = 7 * 2 * 6 * 6 = 504

Therefore, 504 different words are possible where the second letter is a vowel.