Mathematics - Permutation & Combination Question with Solution | TestHub

The letters other than vowels are SHSHNK which can be arranged in $\frac{6!}{2!2!}$ ways

Now in its each case, let the first A be placed in the ${\mathrm{r}}^{\text{th }}$ gap then the number of ways to place the 2nd A will be ( $7-r-1$ ). So, the total number of ways $=\frac{6!}{2!2!}{\sum }_{r=1}^5(6-r)$ $=\frac{6!}{2!2!}\times (5+4+3+2+1)=2700.$