How many 3 letter words can be formed using a b/c d/e if repetition is not allowed?

#32 - #33 page 171

How many selections are there in which either Dolph is chairperson or he is not an officer?
Answer : case that Dolph is the chair plus the case that Dolph is not an officer, and these two cases are mutually disjoint.
How many selections are there in which Ben is either chairperson or treasurer?
Answer : C(5,2)*2! + C(5,2)*2! or 5*4 + 5*4

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(34 - 41 page 171)the letters ABCDE are to be used to form strings of length 3:

How many strings can be formed if repetitions are allowed?
Answer : 53 .Since for each of the three positions , we hae five choices.
Same as before , but repetitions are not allowed.
Answer : 5 * 4 * 3
How many strings begin with A , allowing repetition?
Answer : 52 .
How many strings begin with A if repetitions are not allowed?
Answer: 4 * 3
How many strings do not contain the letter A, allowing repetitions?
Answer : ?? (similar to #34 but take letters only from BCDE)
How many strings do not contain the letter A, if repetitions are not allowed?
Answer : ?? (Similar to #35, but takes letters from BCDE).
How many strings contain letter A, allowing repetition?
Answer : #34 minus #38
How manu strings contain letter A if repetitions are not allowed?
Answer : #35 minus #39

#42- 52 :integers from 5 to 200 , inclusive,

How many numbers are there ?
Answr = 200 - (5-1)
How many are even ?
Answer : half of them
How many are divisible by 5 ?
Answer : [200/5]
How many contain the digit 7 ?
Answer : single-digit case : 1 ; double digit case : 10 + 9 - 1 ; three digit case : (1XY) 10 + 10 -1
How many do not contain 0 ?
Answer : 5 (single digit case) + 9*9 (2-digit case) + 9*9 (1XY case)
How many greater than 101 and do not contain the digit 6 ?
Answer : 1 (case 200) + 9*9 (1XY case) - 2 (case 101 and case 100)
How many have the digits in strictly increasing order?
Answer : 5 (signle-digit case) + (90 - 9-9) /2 (2-digit case) + (3-digit case : 1XY ) (100 - 10)/2
how many consist of distinct digits?
Answer: (single-digit case ) 5 + ??? (double-digit case ) + (three-digit case : 1XY)
How many are of the form xyz, where 0 < x < y and y > z ?
Since x must be 1 , then y > 1, so for y between 2 to 9 , z must less than y. Answer = 2 + 3 + ... + 9 = 44

#10 - 18 p.182 : determine how many strings can be formed by ordering the letters ABCDE subject to the conditions given:

Contains the substring ACE : 3!
Contains the letters ACE together in any order : 3! * (3!)
contains the substrings DB and AE : 3!
contains either the substring AE or the substring EA : 2 * 4!
A appears before D : 5!/2
Contain neither of the substring AB, CD
5! - number of strings contains either AB or CD (or both)
Contains neither of the substring AB, BE :
5! - { 4! + 4! - 3! }
A appears before C and C appears before E :
5! / 3!
Contains either the substring DB or the substring BE :
4! + 4! - 3!

#31-36 refer to a club consisting od six distinct men and seven distinct women :

In how many ways can we select a committee of five persons?
Answer = C(6+7,5)
In how many ways can we select a committee of three men and four women?
Hint : use multiplication principle
In how many ways can we select a committee of four persons that has at least one woman?
Answer = C(7,1) * C(6+ 7-1, 3) or C(6+7,4) - C(6,4)
In how many ways can we select a committee of four persons that has at most one man?
Answer = C(7,4) + C(6,1)*C(7,3)
In how many ways can we select committee of four persons that has persons of both sexes?
Answer = C(13,4) - C(6,4) - C(7,4)
In how many ways can we select a committee of four so that Mabel and Ralph do not serve together?
Answer = C(13,4) - C(11,2)

How many 8-bit strings contain exactly three 0's ?
Answer = 8!/(3!5!) or C(8,3)

How many 8-bit strings contains three 0's in a row and five 1's?
Answer = 6!/5!

#63 - 66 refer to a shipment of 50 microprocessors of which four are defective.

In how many ways can we select a set of four microprocessors?
answer = C(50,4)
In how many ways can we select a set of four non-defective microprocessors?
Answer = C(50-4,4)
In how many ways can we select a set of four microprocessors containing exactly two defective microprocessors?
Hint : How many ways to select 2 defectives and how many ways to select two non-defective?
In how many ways to select a set of four containing at least one defective?
Answer = C(50,4) - C(50-4,4)

If repetition is allowed, we have 4 choices for each letter. Therefore, we can form 4*4*4 = 64 such “words". Good luck!

Solution, The word SIGNATURE has 9 different letters. The number of 3-letter words that can be formed = 3!

The answer is 625 ways.

Without any restrictions on the number of repetitions, we found 216 three-letter words.

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