
How many unique words (may not have any meaning)can be formed by using all the letters from the word NATIONAL so that four vowels are not placed together?
$\left( a \right)$ 18720
$\left( b \right)$ 5040
$\left( c \right)$ 9360
$\left( d \right)$ 10080
Answer
592.5k+ views
Hint: In this particular question use the concept that the number of ways to arrange n different object 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.
Complete step-by-step answer:
Given word:
‘NATIONAL’
As we see that all the letters in the given word are different except 1 and the number of letters are 8.
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 are present in the given word NATIONAL are A, I and O and A, as A is repeated two times
So in the given word there are 4 vowels and 4 consonants.
Now the total number of words which are possible form the letters of a given word NATIONAL = $\dfrac{{8!}}{{2!}}$, as A is repeated one time so divide by 2!
Consider that all the vowels are together so consider four vowels present in the given word as one letter.
So the arrangements of vowels internally = $\dfrac{{4!}}{{2!}}$
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 = \[\left( {5! \times \dfrac{{4!}}{{2!}}} \right)\]
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 = \[\dfrac{{8!}}{{2!}} - \left( {5! \times \dfrac{{4!}}{{2!}}} \right)\]
Now simplify we have,
\[ \Rightarrow \dfrac{{8.7.6.5!}}{{2.1}} - \left( {5! \times \dfrac{{4.3.2.1}}{{2.1}}} \right)\]
\[ \Rightarrow 4.7.6.5! - \left( {12.5!} \right)\]
\[ \Rightarrow 5!\left( {168 - 12} \right) = 120\left( {156} \right) = 18720\]
Hence option (a) is the correct answer.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall that if there are n objects in which p objects are of same type and q objects are of another same type, so the number of ways to arrange them is given as $\dfrac{{n!}}{{p!\left( {q!} \right)}}$
Complete step-by-step answer:
Given word:
‘NATIONAL’
As we see that all the letters in the given word are different except 1 and the number of letters are 8.
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 are present in the given word NATIONAL are A, I and O and A, as A is repeated two times
So in the given word there are 4 vowels and 4 consonants.
Now the total number of words which are possible form the letters of a given word NATIONAL = $\dfrac{{8!}}{{2!}}$, as A is repeated one time so divide by 2!
Consider that all the vowels are together so consider four vowels present in the given word as one letter.
So the arrangements of vowels internally = $\dfrac{{4!}}{{2!}}$
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 = \[\left( {5! \times \dfrac{{4!}}{{2!}}} \right)\]
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 = \[\dfrac{{8!}}{{2!}} - \left( {5! \times \dfrac{{4!}}{{2!}}} \right)\]
Now simplify we have,
\[ \Rightarrow \dfrac{{8.7.6.5!}}{{2.1}} - \left( {5! \times \dfrac{{4.3.2.1}}{{2.1}}} \right)\]
\[ \Rightarrow 4.7.6.5! - \left( {12.5!} \right)\]
\[ \Rightarrow 5!\left( {168 - 12} \right) = 120\left( {156} \right) = 18720\]
Hence option (a) is the correct answer.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall that if there are n objects in which p objects are of same type and q objects are of another same type, so the number of ways to arrange them is given as $\dfrac{{n!}}{{p!\left( {q!} \right)}}$
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw a labelled diagram of the human heart and label class 11 biology CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

