Rework problem 27 from section 1.4 of your text, involving a telephone sales representative. Assume that the representative will continue to make telephone calls until she makes 2 sales, gets 4 tentative commitments, or fails to make a sale. Her log for the day consists of a list of calls, with sale, tentative commitment, or no sale noted for each. How many possible logs are there?

To solve the problem, we need to determine the number of possible logs for the telephone sales representative under the given conditions. The representative will stop making calls when she achieves one of the following outcomes: 1. Makes 2 sales (S) 2. Gets 4 tentative commitments (T) 3. Fails to make a sale (N) Let's denote: - S = Sale - T = Tentative commitment - N = No sale The representative will continue making calls until one of the stopping conditions is met. We can represent the logs as sequences of S, T, and N, with the following constraints: - The sequence must contain exactly 2 S's if the representative stops due to sales. - The sequence must contain exactly 4 T's if the representative stops due to tentative commitments. - The sequence must contain at least one N if the representative stops due to failing to make a sale. ### Case 1: Stopping after making 2 sales (SS) In this case, the log must contain exactly 2 S's. The representative can make any number of T's and N's before the second S. The sequence must end with S. Let \( k \) be the total number of T's and N's before the second S. The total length of the log will be \( k + 2 \) (where 2 is for the two S's). The number of ways to arrange \( k \) T's and N's followed by SS is given by the binomial coefficient: \[ \text{Number of logs} = \binom{k + 1}{k} = k + 1 \] This is because we can place the 2 S's in \( k + 1 \) positions (before the first S, between the S's, and after the second S). The total number of logs for this case is the sum over all possible values of \( k \): \[ \sum_{k=0}^{\infty} (k + 1) = \infty \] ### Case 2: Stopping after getting 4 tentative commitments (TTTT) In this case, the log must contain exactly 4 T's. The representative can make any number of S's and N's before the fourth T. The sequence must end with T. Let \( m \) be the total number of S's and N's before the fourth T. The total length of the log will be \( m + 4 \). The number of ways to arrange \( m \) S's and N's followed by TTTT is given by: \[ \text{Number of logs} = \binom{m + 4 - 1}{4 - 1} = \binom{m + 3}{3} \] This is because we can place the 4 T's in \( m + 4 \) positions. The total number of logs for this case is the sum over all possible values of \( m \): \[ \sum_{m=0}^{\infty} \binom{m + 3}{3} = \infty \] ### Case 3: Stopping after failing to make a sale (N) In this case, the log must contain at least one N. The representative can make any number of S's and T's before the first N. The sequence must end with N. Let \( p \) be the total number of S's and T's before the first N. The total length of the log will be \( p + 1 \). The number of ways to arrange \( p \) S's and T's followed by N is given by: \[ \text{Number of logs} = \binom{p + 1}{1} = p + 1 \] This is because we can place the N in \( p + 1 \) positions. The total number of logs for this case is the sum over all possible values of \( p \): \[ \sum_{p=0}^{\infty} (p + 1) = \infty \] ### Conclusion Since each case leads to an infinite number of possible logs, we conclude that the total number of possible logs for the telephone sales representative is infinite.