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Self Paced 1A Week 5 Advanced Counting
You are correct. The answer for F6 should be $7^4$, not $4^7$.
On F8, we do only care which days someone was born in, but it makes no difference which of them was born on each day. Thus, we only need to select $4$ different days out of the $7$ days of the week., that is, $\binom{7}{4}$.
On F9 we do care about who was born on each day, and since each person was born on a different day, we can use permutations to figure that out. Say you first choose which of the $7$ days Alice was born, then which of the $6$ remaining days Bob was born, and so on. That is, $7\times 6\times 5\times 4$, or $7P4 = \frac{7!}{3!}$.
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