Sol. Model the room as a rectangular solid with dimensions 4m by 4m by 3m, and each ping-pong ball as sphere of diameter 0.038m. The volume of the room is 4 × 4 × 3 = 48 m 3 , while the volume of one ball is .

Therefore, one can fit about ping-pong balls in the room.

As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best packing fraction” is π = 0.74 so that at least 26% of the space will be empty. Therefore, the above estimate reduces to 1.67 × 10 6 × 0.740 ~ 10 6 .