remove particular characters from a variable using bash

There is no need to execute an external program. bash's string manipulation can handle it (also available in ksh93 (where it comes from), zsh and recent versions of mksh, yash and busybox sh (at least)):

$ VERSION='2.3.3'
$ echo "${VERSION//.}"
233

(In those shells' manuals you can generally find this in the parameter expansion section.)

By chronological order:

tr/sed

echo "$VERSION" | tr -d .
echo "$VERSION" | sed 's/\.//g'

csh/tcsh

echo $VERSION:as/.//

POSIX shells:

set -f
IFS=.
set -- $VERSION
IFS=
echo "$*"

ksh93/zsh/mksh/bash/yash (and busybox ash when built with ASH_BASH_COMPAT)

echo "${VERSION//.}"

zsh

echo $VERSION:gs/./

In addition to the successful answers already exists. Same thing can be achieved with tr, with the --delete option.

echo "2.3.3" | tr --delete .
echo "2.3.3" | tr -d .       # for MacOS

Which will output: 233