If the permutations of A, B, C, D, E taken al | vedclass

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(D) To find the rank of $DEBAC$,we list the letters in alphabetical order: $A, B, C, D, E$.$1$. Words starting with $A$: $4! = 24$$2$. Words starting

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$93$

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(D) To find the rank of $DEBAC$,we list the letters in alphabetical order: $A, B, C, D, E$.$1$. Words starting with $A$: $4! = 24$$2$. Words starting with $B$: $4! = 24$$3$. Words starting with $C$: $4! = 24$$4$. Words starting with $DA...$: $3! = 6$$5$. Words starting with $DB...$: $3! = 6$$6$. Words starting with $DC...$: $3! = 6$$7$. Words starting with $DEABC$: $1$$8$. Words starting with $DEACB$: $1$$9$. Words starting with $DEBAC$: $1$Total rank = $24 + 24 + 24 + 6 + 6 + 6 + 1 + 1 + 1 = 93$.Alternatively,using the positional method:$D$ is the $4^{th}$ letter ($3$ letters before it: $A, B, C$): $3 	imes 4! = 72$$E$ is the $4^{th}$ remaining letter ($3$ letters before it: $A, B, C$): $3 	imes 3! = 18$$B$ is the $2^{nd}$ remaining letter ($1$ letter before it: $A$): $1 	imes 2! = 2$$A$ is the $1^{st}$ remaining letter ($0$ letters before it): $0 	imes 1! = 0$$C$ is the $1^{st}$ remaining letter ($0$ letters before it): $0 	imes 0! = 0$Rank = $72 + 18 + 2 + 0 + 0 + 1 = 93$.

If the permutations of $A, B, C, D, E$ taken all together are written down in alphabetical order as in a dictionary and numbered,then the rank of the permutation $DEBAC$ is

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