Question

Find the number of arrangements which can be made out of the letters of the word algebra, without altering the relative positions of vowels and consonants.

Answer 1

$⟹$ Considering $3$ bundles of letters, ${}^{\prime }a{l}^{\prime }$ $,$ ${}^{\prime }ge{b}^{\prime }$ $,$ ${}^{\prime }r{a}^{\prime }.$
Total arrangements of $3$ bundles $=3!$
The bundles can arrange internally in $2!,3!,$ and $2!$ ways.
$\therefore$ Total Number of ways $=\frac{3!\times 2!\times 3!\times 2!}{2!}$
Since ${}^{\prime }{a}^{\prime }$ is twice
$\therefore$ Total no. of ways $=72$