Question

how many different words can be formed out of the letters of the word ALLAHABAD? In how many of them the vowels occupy the even positions?

Answer 1

ALLAHABAD.
$4A^s,2L^s,H,B,D,$ i.e. 9 letters.
Number of words = $\frac{9!}{4!2!}=\frac{9\times 8\times 7\times 6\times 5}{2}$
$=72\times 105=7560$.
There are 4 vowels, and all are alike i.e. $4A^s$
There are 4 even places, i.e. $2^{nd},4^{th},6^{th}$ and $8^{th}$.
There 4 even position can be filled by 4 vowels in $\frac{4!}{4!}=1$ way.
Now we are left with 5 places in which 5 letters out of which $2L^s$ are alike and rest different can be filled in
$\frac{5!}{2!}=5\times 4\times 3=60$ ways.
Hence the total number of words is $60\times 1=60$.