Not exactly what you would call elegant and it assumes that you mean EG8<2
or OR(EG8=2,EG8<3) or EG8=3 but try:

=IF(OR(EG8={1,2,3}),ED8*((((Fees!D8*(C8=1))+(Fees! D18*(C8=2))+(Fees!D28*(C8=3)))*(G8<2))+(((Fees!D9* (C8=1))+(Fees!D19*(C8=2))+(Fees!D29*(C8=3)))*(AND( G8=2,G8<3)))+(((Fees!D10*(C8=1))+(Fees!D20*(C8=2) )+(Fees!D30*(C8=3)))*(AND(G8=3)))),0)

It works because there will only be one number returned out of:

(((Fees!D8*(C8=1))+(Fees!D18*(C8=2))+(Fees!D28*(C8 =3)))*(G8<2))
or
(((Fees!D9*(C8=1))+(Fees!D19*(C8=2))+(Fees!D29*(C8 =3)))*(AND(G8=2,G8<3)))
or
(((Fees!D10*(C8=1))+(Fees!D20*(C8=2))+(Fees!D30*(C 8=3)))*(AND(G8=3)))

with the other two returning zeros and this number is then multplied by ED8

--
HTH

Sandy
In Perth, the ancient capital of Scotland
and the crowning place of kings


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"Stefania" wrote in message
...
Hi!

I'm writing a formula but I can't get my head around it.

The Formula below works:

=IF(EG8<2,IF(C8=1,ED8*Fees!D8,IF(C8=2,ED8*Fees!D18 ,IF(C8=3,ED8*Fees!D28,0))),0)

However I also need to add
IF(EG82<3,IF(C8=1,ED8*Fees!D9,IF(C8=2,ED8*Fees!D1 9,IF(C8=3,ED8*Fees!D29,0))))

and
IF(EG83<6,IF(C8=1,ED8*Fees!D10,IF(C8=2,ED8*Fees!D 20,IF(C8=3,ED8*Fees!D30,0))))

Am I asking too much or can it be done???

Thanks in advance for ANY kind of response!
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