A boy is typing 26 letters of the alphabet. Every letter shows up only once! What is the probability that in at least one of the two words will show up: "open", "abc"
Suppose $\ A = $ the word 'open' shows up and $\ B = $ the word 'abc' shows up.
$$\ P(A) = \frac{{23 \choose 1 } \cdot 22!}{26!} $$ there are 23 ways to arranging a sequence of 4 letters and thats why $\ {23 \choose 1} $ .
$$\ P(B) = \frac{{24 \choose 1}\cdot 23!}{26!} $$
$$\ P(A \cap B) = ?$$ I'm really not sure how to calculate probability of both events happening. and even trying to calculate the opposite ( the probability that both "open" and "abc" doesn't show).