How many different arrangements of the letters in "HYPERGEOMETRIC" begin and end with "E"?
is it just this?: 13!/2! = 3113510400 arrangements that begin and end with e
How many different arrangements of the letters in "HYPERGEOMETRIC" begin and end with "E"?
is it just this?: 13!/2! = 3113510400 arrangements that begin and end with e
Let's fix two $ E'$s as the first and the last letter of the word that will be created:
$E\underbrace{-----\cdots ----}_{12 \text {letter}}E$
Thus the result : $\dfrac {12!}{2!}$ , since there are two $R$'s among the remaining letters(HYPRGOMETRIC) we divide by $2!$ to eliminate overcounting