Answer  Hint: In this question, we first need to find the total number of arrangements possible with the given letters of the word using the permutation formula given by \[{}^{n}{{P}_{r}}\]. Then we need to find the number of words in which two vowels are together but first selecting the two vowels and then arranging all the letters using the formula \[{}^{n}{{C}_{r}}\]. Now, find the number of words in which 3 vowels are together and then subtract 2 vowels together from total words and add 3 vowels together. Complete step by step solution: Now, the given word is TRIANGLE in which there are 3 vowels I, A, E with total letters of 8.Now, let us find the number of words possible with the given 8 letters.As we already know that arrangement of this can be done using the permutations given by the formula\[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]Now, the arrangement of 8 letters can be done in 8! ways. Here, on comparing with the above formula we have\[n=8,r=8\]Now, on substituting the respective values in the formula we get,\[\Rightarrow {}^{8}{{P}_{8}}\]Now, this can be further written as\[\Rightarrow \dfrac{8!}{0!}\]Now, on simplifying this further we get, \[\Rightarrow 8!=40320\]Now, let us find the words in which two vowels are together. Now, the vowels in the given word are I, A, E. Let us select 2 letters out of these and make them a pair and arrange all the letters. Now, making them as a pair we have 7 letters to be arranged and also consider the arrangement of those two vowels among themselves,\[\Rightarrow {}^{3}{{C}_{2}}\times 7!\times 2!\]Now, this can be further written in the simplified form as\[\Rightarrow \dfrac{3!}{2!1!}\times 7!\times 2!\]Now, on further simplification we get,\[\Rightarrow 30240\]Thus, the words in which two vowels are together are 30240Here, we need to find the words in which 3 vowels are together because when we take off the words in which two vowels are together we are also removing the words in which 3 vowels are together so we need to add them backNow, the words in which 3 vowels are together we need to select the vowels and make them as a 1 pair and arrange with the remaining letters\[\Rightarrow {}^{3}{{C}_{3}}\times 6!\times 3!\]Now, this can be further written in the simplified form as\[\Rightarrow \dfrac{3!}{3!0!}\times 6!\times 3!\]Now, on further simplification we get,\[\Rightarrow 4320\]Now, the number of words in which two vowels are not together are given by \[\Rightarrow 40320-30240+4320\]Now, on simplifying this further we get,\[\Rightarrow 14400\]Hence, the correct option is (c).Note:Instead of finding the words in which 2 vowels are together and then subtracting them from total words we can also solve this by first arranging the letters other than vowels and then arrange vowels in between them and then subtract the words in which two vowels come together. Both the methods give the same result.It is important to note that after subtracting the words in which 2 vowels are together from the total number of words we need to add the vowels in which three vowels are together because we need to find the words only in which 2 vowels are not together.
First of all, $\text{TRIANGLE}$ has $8$ distinct letters, $3$ of which are vowels($\text{I, A, E}$) and rest are consonants($\text{T, R, N, G, L}$). While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together. So, I decided to form two 'batteries'. [$\text{V}$ stands for Vowels and $\text{C}$ stands for consonants.] $$\text{V} \text{ C}\text{ V} \text{ C}\text{ V} $$ And, $$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$ If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations. Now, For the first case, $3$ vowels can be arranged in the $3$ spaces required in $3! = 6$ ways From $5$ consonants, $2$ spaces can be filled with consonants in $^5P_2 = 20$ ways One battery, $(8 - 3- 2) = 3$ letters to arrange. Total number of permutations : $6 * 20 * 4! = 2880$. In Second case, From $3$ vowels, $2$ spaces can be filled with vowels in $^3P_2 = 6$ ways From $5$ consonants, $3$ spaces can be filled with consonants in $^5P_3 = 60$ ways. One battery, $(8 - 2- 3) = 3$ letters to arrange. Total number of permutations : $6 * 60 * 4! = 8640$ So, Total number of permutations for the word $\text{TRIANGLE} = 2880 + 8640 = 11520$ Again, My answer is incorrect, according to my textbook. They report $14400$ is the correct answer. So, what did I miss here now? Please elaborate, and I'll be happy with any sort of help. [Seriously, this morning is getting even more hectic for me] Total number of words in ‘TRIANGLE’ = 8 Out of 5 are consonants and 3 are vowels If vowels are not together, taken we have the following arrangement V | C | V | C | V | C | V | C | V | C | V Consonant can be arranged in 5! = 120 ways Vowel occupy 6 places ∴ 3 vowels can be arranged in 6 places = 6P3 = `(6!)/((6 - 3)!)` = `(6!)/(3!)` = 120 ways So, the total arrangement = 120 × 120 = 14400 ways Here, the required arrangement = 14400 ways. |
How many different words can be formed with the letters of the word TRIANGLE having all vowels never together?
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