Question in probability and statistics I'm terrible at questions involving probability, so I'm hoping you can help me!

Suppose there are eight people in a group, each person with a distinct first name.  Each person has a card with their first name written on it.

If the eight cards are randomized and each person selects a card at random, what is the probability that at least one of the first SIX people draws the card with their name on it.

I'm completely baffled by this problem, thanks for the help!
We want to count how many permutations of $\{1,2,\ldots,8\}$ fix at least one of $1,2,3,4,5$ or $6$. Note that the number of permutations that fix one of these points is $7!$, that fix two is $6!$, etc, and that there are $\binom61=6$ elements in $\{1,2,3,4,5,6\}$, $\binom62$ pairs, $\binom63$ triples, etc. Thus by the inclusion-exclusion principle, the number of permutations that fix an element of $\{1,2,3,4,5,6\}$ is 
$$\binom61\cdot 7!-\binom62\cdot 6!+\binom63 5!-\binom64 4!+\binom65 3!-\binom662!=21514$$
Since the total number of permutations of $\{1,2,\ldots,8\}$ is $8!=40320$, the probability that one of the first six people draws a card with their name on it is
$$\frac{21514}{40320}\approx .534$$
We want to count how many permutations of $\{1,2,\ldots,8\}$ fix at least one of $1,2,3,4,5$ or $6$. Note that the number of permutations that fix one of these points is $7!$, that fix two is $6!$, etc, and that there are $\binom61=6$ elements in $\{1,2,3,4,5,6\}$, $\binom62$ pairs, $\binom63$ triples, etc. Thus by the inclusion-exclusion principle, the number of permutations that fix an element of $\{1,2,3,4,5,6\}$ is 
$$\binom61\cdot 7!-\binom62\cdot 6!+\binom63 5!-\binom64 4!+\binom65 3!-\binom662!=21514$$
Since the total number of permutations of $\{1,2,\ldots,8\}$ is $8!=40320$, the probability that one of the first six people draws a card with their name on it is
$$\frac{21514}{40320}\approx .534$$