Run script skipping files - linux

Question

I have a quite simple script I'd like to write just using bash.
Given a folder with 0..N *.XML files; I want to sort those by name and remove N-10 files (leave the last 10 in place).
I've been tinkering with find and tail/head but couldn't figure a way
find /mnt/user/Temporary/1 -name *.xml | tail -n +10 | rm
Please read up. It is about keeping the last 10. If there are 10 or less files, none should be deleted!
EDIT:
As someone closed, but did not repoen the question, here is the solution for those getting here with the same question.
#!/bin/bash
files=()
while IFS= read -r -d $'\0'; do
files+=("$REPLY")
done < <(find . -name *.xml -print0 | sort)
Limit=$((${#files[#]}-10))
count=0
while [ $Limit -gt $count ]; do
rm "${files[count]}"
let count=count+1
done
Maybe some linux "pro" can optimize it or give it some parameters (like limit, path and file pattern) to make it callable anywhere.

Answer 1

EDIT: New answer
#!/usr/bin/env bash
files=$(find *.xml | wc -l)
[ "$files" -lt 10 ] && echo "Files are less than 10..." && exit 1
count=$(($files-10))
for i in $(find *.xml | sort -V); do
[ $count -eq 0 ] && echo "Done" && exit 1
rm $i
((count--))
done
$files stores the number of *.xml in the folder
if the number is less or equal to 10 exit
set a counter that of the number of files to delete
loop through each file in order
if the counter is equal to 0 exit
if not remove the file and increment the counter