This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 16-2. Problem 2. For each integer between 1 and 15 we have 2 choices: we can put it in box 1 or in box 2. Since box 1 has exactly ve integers we must use box 1 ve times and so, we must use box 2 ten times. So we are really arranging ve 1s and ten 2s in row. The requirement that consecutive integers cannot go into box 1 means that there are no consecutive 1s in the arrangement. So let us rst arrange the ve 1s: 1 1 1 1 1 Now we want to distribute the 2s making sure that there are no consecutive 1s. So rst put a 2 into each of the 4 spaces between the 1s. Now, we are left with 6 2s to distribute into the 6 spaces. So there are 6 + 6-1 6 ! ways to do this. 1...
View Full Document
- Spring '11