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6.Permutation and Combination
medium
In a city no two persons have 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
${2^{32}}$
B
${(32)^2} - 1$
C
${2^{32}} - 1$
D
${2^{32 - 1}}$
Solution
(c) We have $32$ places for teeth. For each place we have two choices either there is a tooth or there is no tooth
Therefore the number of ways to fill up these places is ${2^{32}}$. As there is no person without a tooth, the maximum population is ${2^{32}} – 1$.
Standard 11
Mathematics
