2 added 207 characters in body edited Aug 20 '18 at 9:45 Alexei Boulbitch 22.9k2727 silver badges7878 bronze badges Provided each term always has a coefficient (like a[om1,...], c[om1,...] etc.) and may (or may not) have one or several F factors, try this: Select[list, Length[#] == 2 &] (* {c[om1, om2, om3, om4, om5, om6] F[om2]} *)  Edit: taking into account your comment try also this: count[expr_] := Count[expr, _F, Infinity] Select[list, count[#] == 1 &] (* {c[om1, om2, om3, om4, om5, om6] F[om2]} *)  Have fun! Provided each term always has a coefficient (like a[om1,...], c[om1,...] etc.) and may (or may not) have one or several F factors, try this: Select[list, Length[#] == 2 &] (* {c[om1, om2, om3, om4, om5, om6] F[om2]} *)  Have fun! Provided each term always has a coefficient (like a[om1,...], c[om1,...] etc.) and may (or may not) have one or several F factors, try this: Select[list, Length[#] == 2 &] (* {c[om1, om2, om3, om4, om5, om6] F[om2]} *)  Edit: taking into account your comment try also this: count[expr_] := Count[expr, _F, Infinity] Select[list, count[#] == 1 &] (* {c[om1, om2, om3, om4, om5, om6] F[om2]} *)  Have fun! 1 answered Aug 20 '18 at 9:31 Alexei Boulbitch 22.9k2727 silver badges7878 bronze badges Provided each term always has a coefficient (like a[om1,...], c[om1,...] etc.) and may (or may not) have one or several F factors, try this: Select[list, Length[#] == 2 &] (* {c[om1, om2, om3, om4, om5, om6] F[om2]} *)  Have fun!