   Question Answers

# How many arrangements of the letters of the word ‘BENGALI’ can be made? $\left( i \right)$ If all vowels are never together.$\left( {ii} \right)$ If the vowels are to occupy only odd places.$\left( a \right)720,426$$\left( b \right)4320,576$$\left( c \right)822,593$$\left( d \right)644,745$  Verified
129.6k+ views
Hint: In this particular question use the concept that the number of ways to arrange r different objects from n different objects is equal to ${}^n{P_r}$, and the number of ways to arrange n different objects is equal to n!, and use the concept that if we want some letters not together so first find out all the words with these letters together then subtract these words from the total possible words, so use these concepts to reach the solution of the question.

Given word:
‘BENGALI’
As we see that all the letters in the given word are different and the number of letters are 7.
As we all know there are 5 vowels present in the English alphabets which are given as A, E, I, O and U.
Out of these alphabets the number of alphabets present in the given word BENGALI are E, A and I.
So in the given word there are 3 vowels and 4 consonants.
Now the total number of words which are possible form the letters of given word BENGALI = 7!
$\left( i \right)$ If all vowels are never together.
Consider that all the vowels are together so consider three vowels present in the given word as one letter.
So the arrangements of vowels internally = 3!
So there are 5 letters in the word so the number of arrangements = 5!
So the total number of words possible when all the vowels are together = $\left( {5! \times 3!} \right)$
So the total number of words such that no vowel is together is the difference of total number of words from the given word and the total number of words when all the vowels are together.
Therefore, the total number of words such that no vowel is together = $7! - \left( {5! \times 3!} \right)$
Now simplify we have,
$\Rightarrow \left( {7.6.5!} \right) - \left( {6.5!} \right) = 6.5!\left( {7 - 1} \right) = 36\left( {5!} \right) = 36\left( {120} \right) = 4320$ Words.
$\left( {ii} \right)$ If the vowels are to occupy only odd places.
As we see that there are 7 letters in the given word. So there are 4 odd places which is ${1^{st}},{3^{rd}},{5^{th}}{\text{ and }}{{\text{7}}^{th}}$ places.
As there are three vowels so the number of ways to arrange them in 4 odd places = ${}^4{P_3}$.
So three vowels are filled in the odd places, so there are 4 letters remaining and there are 4 places remaining so the number of ways to fill these 4 letters in 4 places = 4!
So the total number of words such that vowels comes in odd places = ${}^4{P_3} \times 4!$
Now as we know that ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ so use this property in the above equation we have,
So the total number of ways such that vowels comes in odd places = $\dfrac{{4!}}{{\left( {4 - 3} \right)!}} \times 4!$
$\Rightarrow \dfrac{{4!}}{{\left( {4 - 3} \right)!}} \times 4! = 4! \times 4! = 24\left( {24} \right) = 576$ Words.
So the total number of words such that vowels come in odd places are 576.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of permutation which is stated above and always recall that in a English alphabets there are 5 vowels which are given as, A, E, I, O, and U, so according to this find out the number of vowels present in the given word.