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read about std::next_permutation

function take_out_char( str, chr ) {
    // TODO: use buffer????
    let result = '';
    for( let n=0; n<from.length; n++ ) {
        if( str[n]!=chr ) {
            result += str[n];
        }
    }
    return result;
}
function get_chars_less_or_equal_to( str, chr ) {
    // TODO: use buffer????
    let result = '';
    for( let n=0; n<str.length; n++ ) {
        if( str[n]<=chr ) {
            result += str[n];
        }
    }
    return result;
}
function factorial( n ) {
    if( n==1 ) return 1;
    return n*factorial(n-1);
}
function count_number_of_strings_between( from, to ) {
    if( from.length!=to.length )
        throw Error("String lengths do not match");
    if( from.length==0 )
        return 0;
    let result = 0;
    // Let's find the number of possible variations with given first character of "to"
    let tmp = get_chars_less_or_equal_to(from,to[0]);
    for( let k=0; k<tmp.length; k++ ) {
       if( tmp[k]==to[0] ) {
           // In this case we need to fix the first char of the "fitting" combination to tmp[k]
           // and count the number of strings between (from without the tmp[k]) and
           // (to without to[0])
           result += count_number_of_strings_between(take_out_char(from,tmp[k]),to.substring(1));
       }
        else {
            // Here tmp[k] is less than to[0].
            // in this case we can make tmp[k] to be the first character and we do not
            // care what the rest of the string is. The number of all possible "rests" is
            // the factorial
            result += factorial(from.length-1);
        }
    }
    return result;
}

let from = 'abc';
let to   = 'ddd';
// Subtract 1 to avoid counting string from itself
console.log(count_number_of_strings_between(from,to)-1);