Question

The total number of arrangements which can be made out of the word ALGEBRA without altering the relative position of vowels and consonants.

Answer 1

Given letters $A,L,G,E,B,R,A=7$ letters.

Vowels $=A,E,A$

Consonants $=L,G,B,R$

Total permutations of the letters $=\frac{7!}{2!}$

Permutations of vowels $=\frac{3!}{2!}=3$

Permutations of consonants $=4!=24.$

Out of $3$ ways only one has the order $A,E,A$ $\xi$ similarly out of $24$ only $1$ has the order $L,G,B,R.$

$\therefore$ By symmitry total possibilities of word with order of vowels $\xi$ consonants fixed is $\frac{1}{3}\times 124\times \frac{7!}{2!}$

Required =$35.$

Hence, the answer is $35.$