Exercise 2. Let A be the set of 26 letters of the alphabet, in lowercase. Let S

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 B S S of all strings in which no letter appears more than (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 once. 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 C.

Explanation / Answer

Part (A)

Set S = {all 6-long letter strings}

The first position of 6-long letter strings can be filled by any one of the 26 letters in 26 ways.

Since the same letter can repeat, the second position of 6-long letter strings can also be filled by any one of the 26 letters in 26 ways. By the same logic, all the other four positions can also be filled in 26 ways each. Thus, the total number of possibilities is:

266 ANSWER

Part (B)

Subset B contains 6-long letter strings in which no letter repeats.

The first position of 6-long letter strings can be filled by any one of the 26 letters in 26 ways.

Since the same letter cannot repeat, the second position of 6-long letter strings can only be filled by any one of the remaining 25 letters in 25 ways. By the same logic, the other four positions can only be filled in 24, 23, 22 and 21 ways respectively. Thus, the total number of possibilities is: 26 x 25 x 24 x 23 x 22 x 21 = (26!)/(20!)

1.66E+08 ANSWER

Part (C)

Subset C contains 6-long letter strings in which no vowel (i.e., a, e, i, o u) appears.

The first position of 6-long letter strings can be filled by any one of the 21 consonants in 21 ways. Since the same letter can repeat, the second position of 6-long letter strings can also be filled by any one of the 21 consonants in 21 ways. By the same logic, the other four positions can also be filled in 21 ways each. Thus, the total number of possibilities is:

216 ANSWER

Part (D)

Subset D contains 6-long letter strings in which fourth and the last (i.e., 6th) letter agree and the same letter can appear in other positions also.

The fourth position of 6-long letter strings can be filled by any one of the 26 letters in 26 ways. Then, the sixth position can be filled by that letter only, i.e., in one way only.

Since the same letter can repeat in other positions, all the other four positions can also be filled in 26 ways each. Thus, the total number of possibilities is: 26 x 1 x 26 x 26 x 26 x 26 =

265 ANSWER

Part (E)

Subset E contains 6-long letter strings in which only fourth and the last (i.e., 6th) letter agree implying that the letters in other positions are all different.

The fourth position of 6-long letter strings can be filled by any one of the 26 letters in 26 ways. Then, the sixth position can be filled by that letter only, i.e., in one way only.

Since the letters in other positions must be different, the other four positions can be filled in 25, 24, 23 and 22 ways respectively. Thus, the total number of possibilities is: 26 x 1 x 25 x 24 x 23 x 22 = (26!)/(21!)

789360 ANSWER

Part (F)

Subset F contains 6-long letter strings in which exactly 3 copies of x appears. These 3 positions can be filled by only x in just one way. Then, the other three positions can be filled in 25, 24 and 23 ways respectively. Thus, the total number of possibilities is:

1 x 25 x 24 x 23 =

13800 ANSWER

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.