- #1

jdinatale

- 155

- 0

## Homework Statement

Part (a). If the cards are distinct, I see an astronomical number of cases possible.

Case 1. All 28 cards in one envelope. There are 9 envelopes. So 9 ways.

Case 2. 27 of the 28 cards in one envelope. There are (28 choose 27) ways of selecting the 27 cards for one envelope. There are 9 envelopes possible. So (28 choose 27)*9. But then there's the one extra card to place in one of the eight other envelopes. There are 8 remaining envelopes. So in case 2, we have (28 choose 27)*9*8 ways.

Case 3 gets REALLY complicated because you can have 26 cards in one envelope and 2 in a second envelope or 26 cards in one envelope, 1 in a second envelope, and 1 in a third envelope.

So...(28 choose 26)*8 + (28 choose 26)*(2 choose 1)*(1 choose 1)*8*7

See why I'm wondering if there's a more efficient way to do this?