Exercise 2. Let A be the set of 26 letters of the alphabet, in lowercase. Let S
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Question
Exercise 2. Let A be the set of 26 letters of the alphabet, in lowercase. Let S be the set of six-long letter strings, in which letters may repeat. Find the size of each of the following subsets. (Your answer can be a number, or a product You may use nCk for "n choose k".) (A) S itself (B) The subset BCS of all strings in which no letter appears more than once (C) The subset C in which the letters a, e, i, o, u do not appear. (D) The subset D in which the fourth and last letters agree. (The same letter may appear elsewhere.) (E) The subset E in which only the fourth and last letters agree. That is, no other letters in two different positions agree (F) The subset F all strings that contain exactly three copies of the letter x.Explanation / Answer
A) We need to find the number of 6-letter strings that can be formed using the 26 alphabets (repetition allowed). Each of the 6 places can contain any alphabet out of 26. Hence, there are 26 possibilities at each place. That is, there are 26 choices for the first letter, 26 choices for the second letter and so on. Now, since these choices are independent of each other, we use the "multiplication principle". Thus the size of 'S' is 26*26*26*26*26*26 = 266
B) There are 26 choices for the first letter. Hence, can choose the first letter in 26 ways. Now, since alphabets cannot repeat, the second letter can be chosen in only 25 ways (The letter that is selected for the first place cannot be chosen for the second place). Similarly, the third letter can now be chosen in only 24 ways, and so on. Also, since these events are independent, we use the "multiplication principle". So, the size of subset 'B' is 26*25*24*23*22*21.
C) Since you have not specified explicitly, I assume that repetition is allowed in this case. This part is similar to the first part. The only difference is that the number of possibilities decrease. Instead of 26, we now have 21 possibilities {26 minus 5 (since a,e,i,o,u cannot be used)}. Hence, using the same logic as in part (A), the size of set 'C' is 216.
D) First of all, we select the letter that appears at the fourth and last place. There are 26 alphabets, hence 26 ways of selecting that letter. Now, we need to select letters for the remaining four places. For each of those four places, there are 26 choices (as repetition is allowed). Since the events are independent, we use multiplication principle. The size of the set is then calculated as follows:
26 (for selecting letter that appears at 4 and last place) *26*26*26*26(for selecting letters at remaining 4 places) = 265
E) First of all, we select the letter that appears at the fourth and last place. There are 26 alphabets, hence 26 ways of selecting that letter. Next, we have to select letters for remaining 4 places. Since repetition is not allowed, we have 25 possibilities for selecting letter at first place, 24 possibilities for selecting letter at second place, and so on. Also, since the events are independent, we use 'multiplication principle'. Thus, the size of E would be
26 (for selecting letter that appears at 4 and last place) *25*24*23*22(for selecting letters at remaining 4 places)
F) First we select 3 positions (out of 6), where we want to place letter 'x'. This can be done in 6C3 ways. Then, we have to fill remaining 3 places. Assuming that repetition is allowed, there are 25 possibilities for each of the 3 places (since, we cannot use 'x' again, but can use other 25 alphabets). Hence, using multiplication principle, the size of subset F would be 6C3*253
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