In a city no two persons have identical set of teeth and there is no peson without a tooth. Also no person has more than 32 teeth. If we disguard the shape and size of tooth and consider only the positioning of the teeth, the maximum population of the city is

Options

A. (a)`2^(32)`
B. (b)`(32)^(2)-1`
C. (c)`2^(32)-1`
D. (d)`2^(31)`

Answer

To solve the problem, we need to determine the maximum population of a city where no two persons have identical sets of teeth, and every person has at least one tooth, with a maximum of 32 teeth.

### Step-by-Step Solution:

1. **Understanding the Problem**:
Each person can have anywhere from 1 to 32 teeth, and the arrangement of these teeth matters. We will consider the presence or absence of a tooth in each of the 32 possible positions.

2. **Calculating the Total Combinations**:
For each of the 32 positions, there are two choices: either a tooth is present or it is absent. Therefore, for 32 positions, the total number of combinations of teeth arrangements is given by:
\[ 2^{32} \]
This represents all possible arrangements of teeth, including the arrangement where no teeth are present.

3. **Excluding the Case with No Teeth**:
Since the problem states that no person can be without a tooth, we need to exclude the arrangement where all 32 positions are empty (i.e., no teeth). This specific arrangement corresponds to:
\[ 2^0 = 1 \]
Thus, we subtract this one arrangement from the total arrangements:
\[ \text{Maximum Population} = 2^{32} - 1 \]

4. **Final Calculation**:
The maximum population of the city, considering that every person must have at least one tooth, is:
\[ 2^{32} - 1 \]

### Conclusion:
The maximum population of the city is \( 2^{32} - 1 \).