Question

The number of arrangement of the word AAKASH in which vowels occupy places which form three consecutive term of an A.P is

Answer 1

Hint: The given word is AAKASH possible A.P’s are \[\left( {1,2,3} \right){\text{ }},{\text{ }}\left( {2,3,4} \right){\text{ }},{\text{ }}\left( {3,4,5} \right){\text{ }},{\text{ }}\left( {4,5,6} \right){\text{ }},{\text{ }}\left( {1,3,5} \right){\text{ }},{\text{ }}\left( {2,4,6} \right)\]After having overview of all the cases we can proceed with the combination and arrangement of each and every cases.

Complete step-by-step answer:
In first four cases vowels are together, so no. of ways of arranging three vowels are $\dfrac{{{}^3{p_3}}}{{3!}}$ ways and the remaining three consonants will be arranged in ${}^3{p_3}$ ways.
And thus the above format is continued for all the first four cases and therefore number of ways for the arrangement of first four cases will be
$
  {}^3{p_3} \times \dfrac{{{}^3{p_3}}}{3} \times 4ways \\
  1 \times 3! \times 4 \\
   = 24 \\
  $
And now for 5th and 6th case number of ways will be
$
  \dfrac{{3!}}{{3!}} \times 3! \times 2 \\
   = 12 \\
  $
As in total 2 cases are there and for both of them no. of ways will be the same. And thus, calculate the number of ways of arrangement for any one and multiply it by 2 as done above
Hence the total number of arrangements will be the sum of all the possible arrangement as stated above.
$
   = 12 + 24 \\
   = 36ways \\
  $
Hence total number of ways for total arrangement from the word AAKASH in which vowels occupy the places which form three consecutive terms of an A.P is 36 ways

Note: Make all the possible cases without any mistake. Apply basic formulas and concepts of combination without any error. And hence at last in order to calculate the total number of ways of arrangements sum up all the possible arrangements.
A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. This means that XYZ is considered a different permutation than ZYX.