Question

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 1

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.