There are two cases.

(a) If the 3 students join the excursion party then the number of combinations will be

P1=C(22,7)

(b) If the 3 students do not join the excursion party.

Then the number of combinations

P2=C(22,10)

If P is the combinations of choosing the excursion party,then

P=P1+P2 = C(22,7)+C(22,10)

=22!/7!8!+22!/10!12!

=(22×21×20×19×18×17×16×15!)/(7×6×5×4×3×2×1×15!) + (22×21×20×19×18×17×16×15×13×12!)/10×9×8×7×6×5×4×3×2×1×12!)

=817190