
Find the number of different words which can be formed from the letters of the word \[LUCKNOW\] when the vowels always occupy even places.
A. \[120\]
B. \[720\]
C. \[400\]
D. None of these
Answer
162.6k+ views
Hint: First, find the number of consonants and vowels present in the given word \[LUCKNOW\]. Then, arrange the letters in even and odd places. In the end, find how many ways the letters can be arranged to get the required answer.
Formula Used: \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Complete step by step solution: The given word is \[LUCKNOW\].
Total number of letters present in the word: 7
Number of consonants: L, C, K, N, and W: 5
Number of vowels: U, and O: 2
Since, in the given word 2 vowels O, and U are present. There are 3 even places.
So, the number of ways of arranging 2 vowels at 3 places are: \[{}^3{P_2}\]
Also, there are 5 consonants present in the word \[LUCKNOW\].
So, the number of ways of arranging 5 consonants at 4 places are: \[{}^5{P_4}\]
Therefore, the number of words formed in which vowels occupy the even places are:
\[{}^3{P_2} \times {}^5{P_4} = \dfrac{{3!}}{{\left( {3 - 2} \right)!}} \times \dfrac{{5!}}{{\left( {5 - 4} \right)!}}\]
\[ \Rightarrow {}^3{P_2} \times {}^5{P_4} = \dfrac{{3!}}{{1!}} \times \dfrac{{5!}}{{1!}}\]
\[ \Rightarrow {}^3{P_2} \times {}^5{P_4} = 6 \times 120\]
\[ \Rightarrow {}^3{P_2} \times {}^5{P_4} = 720\]
Option ‘B’ is correct
Note: Permutation shows the number of possible arrangements of the objects when the order of the arrangement of the objects matters.
While solving this type of problem use the permutation method.
Formula Used: \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Complete step by step solution: The given word is \[LUCKNOW\].
Total number of letters present in the word: 7
Number of consonants: L, C, K, N, and W: 5
Number of vowels: U, and O: 2
Since, in the given word 2 vowels O, and U are present. There are 3 even places.
So, the number of ways of arranging 2 vowels at 3 places are: \[{}^3{P_2}\]
Also, there are 5 consonants present in the word \[LUCKNOW\].
So, the number of ways of arranging 5 consonants at 4 places are: \[{}^5{P_4}\]
Therefore, the number of words formed in which vowels occupy the even places are:
\[{}^3{P_2} \times {}^5{P_4} = \dfrac{{3!}}{{\left( {3 - 2} \right)!}} \times \dfrac{{5!}}{{\left( {5 - 4} \right)!}}\]
\[ \Rightarrow {}^3{P_2} \times {}^5{P_4} = \dfrac{{3!}}{{1!}} \times \dfrac{{5!}}{{1!}}\]
\[ \Rightarrow {}^3{P_2} \times {}^5{P_4} = 6 \times 120\]
\[ \Rightarrow {}^3{P_2} \times {}^5{P_4} = 720\]
Option ‘B’ is correct
Note: Permutation shows the number of possible arrangements of the objects when the order of the arrangement of the objects matters.
While solving this type of problem use the permutation method.
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