
Find the number of different 8-letter arrangement that can be made from the letters of the word DAUGHTER so that
(i) all the vowels occur together
(ii) all the vowels do not occur together
Answer
609.9k+ views
Hint: For solving this problem, we first consider all the possible cases of number of arrangements for the word DAUGHTER. After that, we imply the imposed condition to obtain a possible answer for our problem.
Complete Step-by-Step solution:
In mathematics, a permutation is defined as the arrangement into a sequence or linear order, or if the set is already ordered, a rearrangement of its element. There are basically two types of permutation i.e., with repetition or without repetition.
According to our question, we are given the word “DAUGHTER”.
For solving part (I), we use the concept of repetition not allowed. Now, the total number of letters in the word “DAUGHTER”\[=8\] without any recurring alphabet. Now, the number of vowels in the word DAUGHTER = A, U and E. As per given in the question, all the vowels should occur together. So, we assume AUE as the single letter.
Number of possible ways of rearranging the 3 vowels $={}^{3}{{P}_{3}}=\dfrac{3!}{(3-3)!}$.
Therefore, Arranging the 3 vowels$=3!=6ways.$
Total numbers of letters, we have to arrange = $5+1=6$.
Possible ways of finding numbers of letters $={}^{6}{{P}_{6}}=6!=720$
Total number of arrangements when all vowels occur together$=720\times 6=4320$.
For part (II), we are required to find a number of possible ways when all vowels do not occur together.
So, All the vowels do not occur together = total number of permutations – arrangement when all vowels occur together
Cases of all these vowels do not occur together $=8!-4320=40320-4320$
Therefore, All the vowels do not occur together$=36000$
Hence, the total number of arrangements when all vowels do not occur together is 36000.
Note: The key concept for solving this question is the knowledge of permutations with some imposed restrictions. Students must be careful while applying the restriction to calculate total possible permutations. This knowledge is useful for solving complex problems.
Complete Step-by-Step solution:
In mathematics, a permutation is defined as the arrangement into a sequence or linear order, or if the set is already ordered, a rearrangement of its element. There are basically two types of permutation i.e., with repetition or without repetition.
According to our question, we are given the word “DAUGHTER”.
For solving part (I), we use the concept of repetition not allowed. Now, the total number of letters in the word “DAUGHTER”\[=8\] without any recurring alphabet. Now, the number of vowels in the word DAUGHTER = A, U and E. As per given in the question, all the vowels should occur together. So, we assume AUE as the single letter.
Number of possible ways of rearranging the 3 vowels $={}^{3}{{P}_{3}}=\dfrac{3!}{(3-3)!}$.
Therefore, Arranging the 3 vowels$=3!=6ways.$
Total numbers of letters, we have to arrange = $5+1=6$.
Possible ways of finding numbers of letters $={}^{6}{{P}_{6}}=6!=720$
Total number of arrangements when all vowels occur together$=720\times 6=4320$.
For part (II), we are required to find a number of possible ways when all vowels do not occur together.
So, All the vowels do not occur together = total number of permutations – arrangement when all vowels occur together
Cases of all these vowels do not occur together $=8!-4320=40320-4320$
Therefore, All the vowels do not occur together$=36000$
Hence, the total number of arrangements when all vowels do not occur together is 36000.
Note: The key concept for solving this question is the knowledge of permutations with some imposed restrictions. Students must be careful while applying the restriction to calculate total possible permutations. This knowledge is useful for solving complex problems.
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