In how many ways 10 persons A₁, A₂, A₃, A₄..................A₁₀ can be seated along a row such that
(i) A₁, A₂, A₃ sit together
(ii) A₁, A₂, A₃ sit in a  specified order need not be together
(iii) A₁, A₂, A₃ sit together in a specified order
(iv) A₂, A₃, A₄ sit always after A₁

(i) Ten persons A₁,A₂,........A₁₀ can be arraranged along a row such that
$A_1,A_2,A_3$ sit together is
consider $(A_1,A_2,A_3)$ as one unit $\left(A_1,A_2,A_3\right),A_4,A_5......A_{10}$ can be arranged  in 8! ways and $A_1,A_2,A_3$ can be rearranged in 3! ways
total no.of permutations = 8! x 3!
(ii) $A_1,A_2\dots A_{10}$ are arranged such that $A_1,A_2,A_3$ in specified order.
consider 10 places and arranges $A_4,A_5......A_{10}$ in ${}^{10}{P}_7$ ways .
now $A_1,A_2,A_3$ can be arrange in rearrange 3 places in one ways
total no.of permutations= ${}^{10}{P}_7\times 1$
(iii) $A_1,A_2,A_3$ sit togeter in specified order 
consider $(A_1,A_2,A_3)$ as one unit arrange $\left(A_1,A_2,A_3\right),A_4,A_5......A_{10}$ in 8! ways 
$A_1,A_2,A_3$ are in one specified order 
no.of permutations =8! x 1 = 8!
(iv)  $A_2,A_3,A_4$ sit always after A1 consider 10 places arrange  A₅,A₆ ............A₁₀ in ${}^{10}{P}_6$ ways
In the rearranging 4 places first place is gives ti A₁ then $A_2,A_3,A_4$ can be arranged in 3! ways
no.of permutations =   ${}^{10}{P}_6\times 3!$