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You can start with the combination $AEgrdn$. The consonants can be arranged in $4!=24$ ways. Now the letter $E$ can be put on the positions $2,3,4,5,6$. Thus in total we have $5\cdot 24=120$ with A at position 1 and E behind. Next $A$ is put on position 2: $gAErdn$. Again the consonants can be arranged in $24$ ways. And $E$ can be put on the positions $3,4,5,6$. Therefore in total we have $4\cdot 24=96$ with $A$ at position $2$ and $E$ behind. Proceed in this manner till $A$ at position $5$ and $E$ at postion $6$. |