Solution to Problem 13

The 12 trees can be planted in 12! orders. Let k be the number of orders in which no two birch trees are adjacent to one another. The probability we need is $\frac{k}{12!}$. To find k, we will count the number of patterns

$$\begin{array}{ccccccccccccccc} \underline{\ \ \ \ }& N &\unde... ...&\underline{\ \ \ \ }\\ 1 &&2&&3&&4&&5&&6&&7&&8\\ \end{array}$$

where the 7 N's denote nonbirch (i.e., maple or oak) trees, and slots 1 through 8 are to be occupied by birch trees, at most one in each slot. There are 7! orders for the nonbirch trees, and for each ordering of them there are $8\cdot 7\cdot 6\cdot 5\cdot 4$ ways to placethe birch trees. Thus, we find that $k=(7!)\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4$ and

$$\frac{m}{n}=\frac{(7!)\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}{12!}= \frac{7}{99} \quad\mbox{ and }\quad m+n=106.$$

Note:    We have assumed in this solution that each tree is distinguishable. The problem can also be interpreted to mean that trees are distinguishable if and only if they are of different species. In that case, the calculations (i.e., the numerator and the denominator) are different, but the probability turns out to be the same.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Tue Nov 21 15:23:13 CST 2000