#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
- 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
- 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
- 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!
- 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]
Answer = 8!/[3!5!] or C[8,3]
Answer = 6!/5!
- 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]
How many 3
If repetition is allowed, we have 4 choices for each letter. Therefore, we can form 4*4*4 = 64 such “words". Good luck!
How many 3
Solution, The word SIGNATURE has 9 different letters. The number of 3-letter words that can be formed = 3!
How many ways 4 letters can be formed from a b/c d/e and f if repetition is not allowed?
The answer is 625 ways.
How many three letter combinations can be made from a b/c d/e and f without repeating a letter?
Without any restrictions on the number of repetitions, we found 216 three-letter words.