## Permutation and Combination - Tips

Permutation - arrangement - example
Question 1 :   A family of 4 brothers and 3 sisters is to be arranged in a row for a photograph. In how many ways can they be seated if all the sisters are together?
Answer:    Let B1,B2,B3,B4 denote the brothers and S1,S2,S3 denote the sisters. Since the sisters are to be seated together for a photograph, consider all the sisters as one unit or entity. Then B1,B2,B3,B4,S can be arranged to sit in 5! ways. The sisters can be arranged among themselves in 3! ways. Since the two events are independent, the total number of arrangements = 5!.3! = 720 ways.

Question 2:   In how many ways can a consonant and a vowel be chosen out of the letters in the word COURAGE?
Answer:    There are three consonants (C, R, G) and four vowels (A, E, O, U) in the word COURAGE.
With the consonants, we may choose any one of the 4 vowels. It can be done in 4 ways. There are three consonants.
\The total number of ways will be 4 x 3 = 12.

Question 3:   How many arrangements can be made out of the letters of the word DRAUGHT, the vowels never being separated?
Answer:    There are 7 letters in the word DRAUGHT, the two vowels are A and U. Since, the vowels are not to be separated, AU can be considered as one entity. Therefore, the number of letters will be 6 instead of 7. The permutations will be P(6,6) = 6! ways.
But the two vowels A and U can be arranged in two ways, i.e. AU and UA.
\The required number of arrangements = 2!.6! = 1440 ways.

Question 4:( Out of SPM Syllibus)   Find the number of arrangements that can be made out of the letters i) ASSASSINATION ii) GANESHPURI.
Answer:    i) The word ASSASSINATION consists of
A's = 3, S's = 4, I's = 2,
N's = 2, T's = 1, O's = 1
The total number of letters is 13 letters.

ii) The word GANESHPURI consists of 10 distinct letters.
The number of permutations is 10!.
Question 5 : ( out of SPM Syllibus)   How many different arrangements can be made out of the letters in the expression a3b2c4, when written at full length?
Answer:    There are 3 + 2 + 4 = 9 letters.
Category:

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Is this a premium wp theme?
The text passage contains a series of permutation and combination problems along with their solutions. Each question is followed by a detailed answer explaining the steps to solve the problem.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
## Permutation and Combination - Tips

Permutation - arrangement - example
Question 1 :   A family of 4 brothers and 3 sisters is to be arranged in a row for a photograph. In how many ways can they be seated if all the sisters are together?
Answer:    Let B1,B2,B3,B4 denote the brothers and S1,S2,S3 denote the sisters. Since the sisters are to be seated together for a photograph, consider all the sisters as one unit or entity. Then B1,B2,B3,B4,S can be arranged to sit in 5! ways. The sisters can be arranged among themselves in 3! ways. Since the two events are independent, the total number of arrangements = 5!.3! = 720 ways.

Question 2:   In how many ways can a consonant and a vowel be chosen out of the letters in the word COURAGE?
Answer:    There are three consonants (C, R, G) and four vowels (A, E, O, U) in the word COURAGE.
With the consonants, we may choose any one of the 4 vowels. It can be done in 4 ways. There are three consonants.
\The total number of ways will be 4 x 3 = 12.

Question 3:   How many arrangements can be made out of the letters of the word DRAUGHT, the vowels never being separated?
Answer:    There are 7 letters in the word DRAUGHT, the two vowels are A and U. Since, the vowels are not to be separated, AU can be considered as one entity. Therefore, the number of letters will be 6 instead of 7. The permutations will be P(6,6) = 6! ways.
But the two vowels A and U can be arranged in two ways, i.e. AU and UA.
\The required number of arrangements = 2!.6! = 1440 ways.

Question 4:( Out of SPM Syllibus)   Find the number of arrangements that can be made out of the letters i) ASSASSINATION ii) GANESHPURI.
Answer:    i) The word ASSASSINATION consists of
A's = 3, S's = 4, I's = 2,
N's = 2, T's = 1, O's = 1
The total number of letters is 13 letters.

ii) The word GANESHPURI consists of 10 distinct letters.
The number of permutations is 10!.
Question 5 : ( out of SPM Syllibus)   How many different arrangements can be made out of the letters in the expression a3b2c4, when written at full length?
Answer:    There are 3 + 2 + 4 = 9 letters.
Category:

Anonymous said...

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-------

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[url=http://pozycjonowanie.lagata.pl]Pozycjonowanie[/url]

[url=http://brum-brum.pl/zdrowie,i,uroda/opony,s,12286/]opony[/url]

Anonymous said...

Is this a premium wp theme?
user
The text passage contains a series of permutation and combination problems along with their solutions. Each question is followed by a detailed answer explaining the steps to solve the problem.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS