Number of ways 7 letters ABCDEFG can be arranged so that A is in 1st position or G is in last position?
closed as unclear what you're asking by user91500, Behrouz Maleki, Leucippus, R_D, barak manos Nov 28 '16 at 6:59
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
Let $A$ be the event that the arrangement starts with the letter $A$. Then $n(A) = 6!$.
Let $B$ be the event that the arrangement ends with the letter $G$. Then $n(G) = 6!$.
Let $C$ be the event that the arrangement starts with the letter $A$ and ends with the letter $G$. Then $n(C) = 5!$. Basically $C = A \cap B$.
Hence, the required answer ($n(A \cup B)$) is $n(A) + n(B) - n(C) = 720 + 720 - 120 = 1320$