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\( 2n = 34 \), so \( n = 17 \).
#### 17, 18Question: A palynologist is analyzing the distribution of pollen samples from 5 different plant species labeled A through E. If each sample must be analyzed exactly once and the order of analysis matters, how many distinct sequences are possible if species A must be analyzed before species B?
To solve this, we begin by considering all permutations of the 5 species. Since each species is analyzed exactly once and order matters, the total number of unrestricted permutations is:
Now, we impose the condition that species A must be analyzed before species B. In any permutation of the 5 species, for every pair A and B, A appears either before B or after B, with equal likelihood. Thus, exactly half of all permutations satisfy the condition that A comes before B.
So, the number of valid sequences is:
\frac{5!}{2} = \frac{120}{2} = 60
\boxed{60}
Question: A wildlife conservation genomics researcher is tracking 6 distinct mountain gorillas for genetic sampling. If the researcher plans to randomly select 3 gorillas to tag on Monday and the remaining 3 on Tuesday, what is the probability that two specific gorillas, Gorilla X and Gorilla Y, are both tagged on the same day?
We begin by computing the total number of ways to divide 6 distinct gorillas into two groups of 3, where the order of the days (Monday and Tuesday) matters. This is simply:
Now, we count the number of favorable outcomes where Gorilla X and Gorilla Y are both on the same day. There are two cases: