The question is: "**What is the minimum number of leaves in the decision tree for a comparison sort? Use this observation to derive a lower bound on the number of comparisons performed by a comparison sort in the worst case**."

In a comparison sort, the decision tree represents all the possible comparisons made between elements during the sorting process. Each leaf node in the decision tree represents a possible permutation of the input elements after sorting.

For 𝑛*n* distinct elements, there are 𝑛!* *possible permutations. Each permutation corresponds to a unique leaf in the decision tree. The minimum number of leaves in the decision tree for a comparison sort is equal to the number of permutations of 𝑛*n* distinct elements, which is 𝑛!.

Now, let's derive a lower bound on the number of comparisons performed by a comparison sort in the worst case based on this observation.

Suppose a comparison sort algorithm constructs a decision tree to sort 𝑛*n* distinct elements. In the worst*-*case scenario, the algorithm must traverse the entire decision tree to find the correct permutation (i.e., the sorted order). Since each leaf in the decision tree represents a unique permutation of the input elements, the worst-case number of comparisons performed by the algorithm is at least equal to the depth of the decision tree.

In a complete binary tree, the number of leaves is 2^d, where *d* is the depth of the tree. So, for *n*! leaves, we have:

Taking the logarithm base 2 of both sides:

Using Stirling's approximation, we have,

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