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Number of ways 7 letters ABCDEFG can be arranged so that A is in 1st position or G is in last position?

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closed as unclear what you're asking by user91500, Behrouz Maleki, Leucippus, R_D, barak manos Nov 28 '16 at 6:59

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    $\begingroup$ I don't see a question, only a question mark. $\endgroup$ – Shraddheya Shendre Nov 28 '16 at 5:31
  • $\begingroup$ The number of ways 7 letters can be arranged minus the number of ways 7 letters can be arranged where neither the A is first nor the G is last. $\endgroup$ – fleablood Nov 28 '16 at 5:39
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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 $n(A) + n(B) - n(C) = 720 + 720 - 120 = 1320$

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