How many ways can letter of the word fraction be arranged so that two vowels are together?

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Solution

The correct option is C14400 The explanation for the correct answer.Find the number of ways such that no two vowels occur together.Given: TRIANGLE (adsbygoogle = window.adsbygoogle || []).push({}); The total number of letters in the word triangle is 8.The total number of words that can be formed =8!=40320.The total number of words in which two vowels occur together =C2×7!×2!=30240 3.The number of ways where all vowels occur together =C3×6!×3!=4320 3.Therefore required number of words =40320-30240+4320=14400.Hence option (C) is the correct answer.

The number of ways in which the letters of the word TRIANGLE can be arranged such that two vowels not occur together is (a) 1200(b) 2400(c) 14400(d) none of these

Answer

Verified

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 30240
Here, 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 back
Now, 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.

total ways in which the letters of the word FRACTION can be arranged = 8! = 40320
vowels : A I O ( let us consider them as 1 letter instead of 3 letters and let this new letter be * )
consonants : F R C T N
now the new word thus formed will be : * F R C T N
the letters of this new word can be arranged in 6! ways. not only that * can arrange itself in 3! ways 
so the total ways in which all the vowels are together = 6! x 3! = 4320
so total words that can be formed so that no two vowels are together = 40320 - 4320 = 36000

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