bash removing part of a file name

Without using an other tools like rename or sed and sticking strictly to bash alone:

for f in CombinedReports_LLL-*.csv
do
  newName=${f/LLL-*\(/LLL-(}
  mv -i "$f" "$newName"
done

for f in CombinedReports_LLL-* ; do
    b=${f:0:20}${f:34:500}
    mv "$f" "$b"
done

You can try line by line on shell:

f="CombinedReports_LLL-20140211144020(Untitled_11).csv"
b=${f:0:20}${f:34:500}
echo $b

rename is part of the perl package. It renames files according to perl-style regular expressions. To remove the dates from your file names:

rename 's/[0-9]{14}//' CombinedReports_LLL-*.csv

If rename is not available, sed+shell can be used:

for fname in Combined*.csv ; do mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')" ; done

The above loops over each of your files. For each file, it performs a mv command: mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')" where, in this case, sed is able to use the same regular expression as the rename command above. s/[0-9]{14}// tells sed to look for 14 digits in a row and replace them with an empty string.