Question Number 33230 by Rio Mike last updated on 13/Apr/18
$${how}\:{many}\:{ways}\:{can}\:{the}\:{word}\: \\ $$$$\:\:\:\:{MOGGOMBABA}\: \\ $$$${be}\:{arranged}\:{hence}\:{find}\:{S}_{\infty} \:{of}\:{a}\: \\ $$$${sequence}\:{a},{ar},{ar}^{\mathrm{2}} \:{in}\:{integral}\: \\ $$$$\:\:\mathrm{1},\sqrt{\mathrm{2}}\:,\:\mathrm{2},... \\ $$
Answered by Joel578 last updated on 14/Apr/18
$$\left(\mathrm{1}\right)\:\mathrm{Word}\:``{MOGGOMBABA}'' \\ $$$$\:\:\:\:\:\:\:\:\mathrm{can}\:\mathrm{be}\:\mathrm{arranged}\:\mathrm{in}\:=\:\frac{\mathrm{10}!}{\mathrm{2}!\:\mathrm{2}!\:\mathrm{2}!\:\mathrm{2}!\:\mathrm{2}!}\:=\:\mathrm{362880}\:\mathrm{ways} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{1},\:\sqrt{\mathrm{2}},\:\mathrm{2},\:\mathrm{2}\sqrt{\mathrm{2}},\:\mathrm{4},\:... \\ $$$$\:\:\:\:\:\:\:\:{a}\:=\:\mathrm{1},\:{r}\:=\:\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Since}\:{r}\:>\:\mathrm{1}\:\mathrm{hence}\:{S}_{\infty} \:\mathrm{diverge} \\ $$