Con,

If the digits were all distinct you could line them up in 7x6x5x4x3x2x1 (7 factorial, written 7!) ways but they are not all distinct so you are counting some lineups more than once. In particular there are 3 8's, now 3 distinct digits could be lined up in 3! = 3x2x1 ways so we have actually counted some lineups 6 times. Similarly, because of the 2's we have counted some lineups 2! = 2x1 times. The net result is that you have 7!/(3!2!) possibilities all told.

Penny