Question

In a city no person has identical set of teeth and there is no person without a tooth. Also, no person has more than 32 teeth. If we disregard the shape and size of tooth and consider only the positioning of the teeth, then the maximum population of the city is:
\[
  A.{\text{ }}{2^{32}} \\
  B.{\text{ }}{2^{32}} - 1 \\
  C.{\text{ }}{2^{32}} + 1 \\
  D.{\text{ None of these}} \\
 \]

Answer 1

Hint: In order to solve the given problem use the concept of permutation and combination along with the fundamental principle of counting. Use the basic concept of availability and unavailability of teeth for each position rather than counting directly person wise to solve the problem and finally remove the impossible case to get the final answer.

Complete step-by-step answer:
We have to find out the number of persons in the town given that everyone has different arrangement of the teeth.
We know that there are 32 places for teeth in the mouth.
For each 32 places in the mouth there are two conditions.
Teeth is present
Teeth is absent
For one tooth there are two possibilities.
For two teeth no of possibilities $ = 2 \times 2 = {2^2}$
For three teeth no of possibilities $ = 2 \times 2 \times 2 = {2^3}$
Similarly proceeding we have-
-
For 32 teeth no of possibilities $ = {2^{32}}$
Therefore, total choice for 32 teeth $ = {2^{32}}$
But this includes one case where there are no teeth at all the 32 positions.
So, we have to subtract that case
Therefore, the required number of ways is $ = {2^{32}} - 1$
Hence, the maximum population of the city is $ = {2^{32}} - 1$

Note: The above problem can also be solved by using the formula of permutation and combination, but it has been solved in the basic ways to make it more understandable. In order to solve such problems, students must keep in mind the impractical and impossible cases and must remove or subtract them before gaining a final answer.