The formula that these isomers contain no rings or double bonds. The isomers must be alcohols and ethers.
Let's start by drawing the alcohols.
Start with four carbons in a row and put an OH group in every possible position.
This gives us
1. CH3CH2CH2CH2OH, butan 1-ol
and
2. CH3CH2CH(OH)CH3, butan-2-ol
Now use a 3-carbon chain with a CH3 on the middle carbon.
This gives
(CH3)2CHCH2OH, 2-methylpropan-1-ol
and
(CH3)3COH, 2-methylpropan-2-ol.
Now for the ethers.
Let's start with five atoms in a row.
This gives
CH3CH2CH2OCH3, 1-methoxypropane
and
CH3CH2OCH2CH3 , ethoxyethane
Finally, the only choice for a branched-chain ether is
(CH3)2CHOCH3 , 2-methoxpropane.
And there are your seven isomers of C4H10O.