Since there is room for 2+4+5=11 passengers but only 10 are going,
There are three cases in which a different car carries 1 person
less than its full capacity.
Case 1. The 2-passenger car carries only 1 person, the other two cars
are full.
10 choose 1, 9 choose 4, 5 choose 5
(10C1)(9C4)(5C5)
Case 1. The 4-passenger car carries only 3 people, the other two cars
are full.
10 choose 2, 8 choose 3, 5 choose 5
(10C2)(8C3)(5C5)
Case 3. The 5-passenger car carries only 4 people, the other two cars
are full.
10 choose 2, 8 choose 4, 4 choose 4
(10C2)(8C4)(4C4)
Grand total: (10C1)(9C4)(5C5) + (10C2)(8C3)(5C5) + (10C2)(8C4)(4C4) =
(10)(126)(1) + (45)(56)(1) + (45)(70)(1) = 1260 + 2520 + 3150 = 6930
Edwin