In how many ways 4 persons can occupy 10 chairs in a row ,if no two sit on adjacent chairs.

Given 4 persons and 10 chairs.
First place 10-4=6 chairs(C) leaving a gap(G) among each chair as follows:

As a result we have 7 gaps, these 7 gaps can filled with 4 persons as ${}^7{C}_4$ ways and also multiplying by the number of ways to arrange the persons themselves to get required arrangements.
So, required number of ways${=}^7{C}_4(4!)=\frac{7!}{3!\times 4!}(4!)$
$=\frac{7\times 6\times 5\times 4\times 3!}{3!\times 4!}(4!)$
$=840$ ways.