wouldn't it be 5C2*5! + 5C1*5!+5!=1920
i dont get why there are two ways to remove an A (you say C(2,1) ), and why do you divide by 2! when you 'remove 2 of "LGEBR" and jumble the rest to make the 5 letter word', i mean you are going to have two A's in this case in the set of 5.
just wondering..