The number of different words which can be formed from the letters of the word $$LUCKNOW$$ when the vowels always occupy even places is
A.
120
B.
720
C.
400
D.
None of these
Answer :
720
Solution :
In $$LUCKNOW,$$ number of letters $$= 7$$ (all distinct), vowels $$\left( {U,O} \right) = 2,$$ consonants $$\left( {L,C,K,N,W} \right) = 5$$
There are 7 letters so 7 places are required. Number of even places are 3 and vowels are 2, so 2 vowels can be placed in $$^3{C_2} \cdot 2! = 6$$ and 5 consonants can be placed in remaining places in $$5!$$ ways.
Hence, number of words $$ = 6 \cdot 5! = 720.$$
Releted MCQ Question on Algebra >> Permutation and Combination
Releted Question 1
$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$ and $$^n{C_{r + 1}} = 126,$$ then $$r$$ is:
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are
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