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How many arrangements can be made out of the letters of the word INTERFERENCE so that no two consonants are together?the have given 240 as anwer,but this is not looks right,how can be this arrangement possible.plz explain?
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How many arrangements can be made out of the letters of the word INTER...
Since the options are not given, I am giving a general answer.
There are 5 vowels in the 'Interference', i.e, - I, e, e,
There are 7 consonants in the word - N, T, R, F, R, N, C.
So if no two consonants should be together, then it should be placed in between the vowels. So, only 6 consonants can be placed that way. But there are 7 consonants in the given word. So, there should be at least 1 instance where 2 consonants will come together. Else, there will be no words formed when two consonants should not be together.
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How many arrangements can be made out of the letters of the word INTER...
Problem Statement: How many arrangements can be made out of the letters of the word INTERFERENCE so that no two consonants are together?
To solve this problem, we can use the concept of permutations and combinations. Let's break down the solution into steps:
Step 1: Identify the total number of letters in the word INTERFERENCE.
The word INTERFERENCE consists of 12 letters.
Step 2: Identify the number of consonants in the word INTERFERENCE.
The consonants in the word INTERFERENCE are N, T, R, F, R, N, C. So there are 7 consonants.
Step 3: Identify the number of vowels in the word INTERFERENCE.
The vowels in the word INTERFERENCE are I, E, E, E. So there are 4 vowels.
Step 4: Identify the number of spaces required to separate the consonants.
Since we want no two consonants to be together, we need to determine the number of spaces required to separate the consonants. We have 7 consonants, so we need 6 spaces.
Step 5: Arrange the vowels in the spaces.
Since the vowels can be arranged in any order, we need to find the number of ways to arrange the vowels in the spaces we identified in Step 4. We have 4 vowels and 6 spaces, so the number of arrangements of the vowels is 6P4 = 6! / (6-4)! = 6! / 2! = 720 / 2 = 360.
Step 6: Arrange the consonants in the remaining spaces.
Now that we have arranged the vowels, we can arrange the consonants in the remaining spaces. We have 7 consonants and 7 spaces (including the spaces at the ends), so the number of arrangements of the consonants is 7P7 = 7! = 5040.
Step 7: Multiply the number of arrangements of vowels and consonants.
To find the total number of arrangements, we need to multiply the number of arrangements of vowels and consonants. 360 (vowels) * 5040 (consonants) = 1,814,400.
Therefore, there are 1,814,400 possible arrangements of the letters of the word INTERFERENCE such that no two consonants are together.
The given answer of 240 is incorrect.
To solve this problem, we can use the concept of permutations and combinations. Let's break down the solution into steps:
Step 1: Identify the total number of letters in the word INTERFERENCE.
The word INTERFERENCE consists of 12 letters.
Step 2: Identify the number of consonants in the word INTERFERENCE.
The consonants in the word INTERFERENCE are N, T, R, F, R, N, C. So there are 7 consonants.
Step 3: Identify the number of vowels in the word INTERFERENCE.
The vowels in the word INTERFERENCE are I, E, E, E. So there are 4 vowels.
Step 4: Identify the number of spaces required to separate the consonants.
Since we want no two consonants to be together, we need to determine the number of spaces required to separate the consonants. We have 7 consonants, so we need 6 spaces.
Step 5: Arrange the vowels in the spaces.
Since the vowels can be arranged in any order, we need to find the number of ways to arrange the vowels in the spaces we identified in Step 4. We have 4 vowels and 6 spaces, so the number of arrangements of the vowels is 6P4 = 6! / (6-4)! = 6! / 2! = 720 / 2 = 360.
Step 6: Arrange the consonants in the remaining spaces.
Now that we have arranged the vowels, we can arrange the consonants in the remaining spaces. We have 7 consonants and 7 spaces (including the spaces at the ends), so the number of arrangements of the consonants is 7P7 = 7! = 5040.
Step 7: Multiply the number of arrangements of vowels and consonants.
To find the total number of arrangements, we need to multiply the number of arrangements of vowels and consonants. 360 (vowels) * 5040 (consonants) = 1,814,400.
Therefore, there are 1,814,400 possible arrangements of the letters of the word INTERFERENCE such that no two consonants are together.
The given answer of 240 is incorrect.
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How many arrangements can be made out of the letters of the word INTERFERENCE so that no two consonants are together?the have given 240 as anwer,but this is not looks right,how can be this arrangement possible.plz explain?
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How many arrangements can be made out of the letters of the word INTERFERENCE so that no two consonants are together?the have given 240 as anwer,but this is not looks right,how can be this arrangement possible.plz explain? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about How many arrangements can be made out of the letters of the word INTERFERENCE so that no two consonants are together?the have given 240 as anwer,but this is not looks right,how can be this arrangement possible.plz explain? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How many arrangements can be made out of the letters of the word INTERFERENCE so that no two consonants are together?the have given 240 as anwer,but this is not looks right,how can be this arrangement possible.plz explain?.
How many arrangements can be made out of the letters of the word INTERFERENCE so that no two consonants are together?the have given 240 as anwer,but this is not looks right,how can be this arrangement possible.plz explain? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about How many arrangements can be made out of the letters of the word INTERFERENCE so that no two consonants are together?the have given 240 as anwer,but this is not looks right,how can be this arrangement possible.plz explain? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How many arrangements can be made out of the letters of the word INTERFERENCE so that no two consonants are together?the have given 240 as anwer,but this is not looks right,how can be this arrangement possible.plz explain?.
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