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Creating words with every letter combination possible

I'm trying to make program in javascript that makes words of given letters (e.g. given letters: "a,b,c", words: "aaaa", "aaab"... and so on.
const random = (length = 6) => {
// Declare all characters
let chars = 'aąbcćdeęfghijklłmnńoópqrsśtuvwxyzźżAĄBCĆDEĘFGHIJKLŁMNŃOÓPQRSŚTUVWXYZŹŻ';
// Pick characters randomly
let str = '';
for (let i = 0; i < length; i++) {
str += chars.charAt(i);
}
return str;
};
Is there any way to do it?
Thanks in advance

If what you're looking for is every possible combination of letters, randomness isn't the way to do this.
Try taking a more methodical approach
combinations = []
function allPossibleCombinations(lettersRemaining, lettersSoFar){
if (lettersRemaining.length == 0){
combinations.push(lettersSoFar);
}else{
for (var i = 0; i < lettersRemaining.length; i++){
allPossibleCombinations(lettersSoFar + lettersRemaining.charAt(i), lettersRemaining.slice(0, i) + lettersRemaining.slice(i + 1))
}
}
}
The code is untested, but this general recursive approach should work

Duplicate a character depending on its position in a string

I would like to duplicate every single letter in my string and uppercasing the first letter.
Like this case:
accum("abcd") -> "A-Bb-Ccc-Dddd".
However, it alters the first letter of the string. I think I should add another iterator called "j". But I don't know how to do it.
Precisely, the only task remaining in my code is to move on to the next letter while saving the changes made for the first letter.
function accum(s) {
var i = 0;
while ( i<s.length){
for (var j =i; j<i ; j++) {
s=s[j].toUpperCase()+s[j].repeat(j)+"-";
i+=1;
}
}
return s.slice(0,s.length-1);
}

Try this:
function accum(s) {
let newString = '';
for(let i = 0; i < s.length; i++) {
newString += s[i].toUpperCase() + s[i].repeat(i) + "-";
}
return newString.slice(0, newString.length - 1);
}
I guess you don't need two repetition loops at all (either you can keep the for or the while, i kept the for).
Your fundamental mistake was in this line: s=s[j].toUpperCase()+s[j].repeat(j)+"-"; where you replaced s with the new string instead of concatenating it (s += instead of s = ). Which would be wrong anyway because you are replacing the original string. You need another empty string to keep the changes separated from the original one.

Do this:
function accum(s) {
accumStr = '';
for (var i=0; i < s.length; i++) {
for (var j = 0; j <= i; j++) {
accumStr += j !== 0 ? s[i] : s[i].toUpperCase();
}
}
return accumStr;
}
console.log(accum('abcd')) //ABbCccDddd

Try this:
function accum(s) {
let strArr = s.split('');
let res = [];
for (let i in strArr) {
res.push(strArr[i].repeat(parseInt(i)+1));
res[i] = res[i].charAt(0).toUpperCase() + res[i].slice(1);
}
return res.join('-');
}
console.log(accum('abcd'))

try to use reduce method
const accum = (str) => {
return [...str].reduce(
(acc, item, index, arr) =>
acc +
item.toUpperCase() +
item.repeat(index) +
(index < arr.length - 1 ? "-" : ""),
""
);
};
console.log(accum("abcd")); //A-Bb-Ccc-Dddd

How to get string from odd's array ?

I need help... I'm learning JavaScript, and it seems easy, but I may just be overlooking... everything... My problem is I need to return a string of all the even numbers between 0 and num inclusive; ex 7 gets you 0246, etc. I've gotten:
function stringOfEvens(num) {
for (let i = 0; i <= num.length; ++i) {
if (i % 2 === 0 ); {
return i;
}
}
}
I know the whole outlook is to run a for loop that goes from 0 to the number in question appending each time and if that number % 2 = 0, return that number, but something is a miss... I may even be overthinking and approaching this whole thing wrong... It is figuratively driving me mad...

function theJob(limit) {
var res = ''; //initialize as a string so that the other numbers will be appended instead of added
for (i = 0; i <= limit; i += 2) { // step increase by 2 to skip odd numbers
res += i;
}
return res; // returning the resulting string
}
console.log(theJob(10));

You can do this without using the modular function, by simply starting the loop at zero and incrementing by 2 each time the loop iterates...
function stringOfEvens(num) {
var result = "";
for (let i = 0; i <= num; i += 2) {
result += i; // append this even number to the result string
}
return result;
}
console.log(stringOfEvens(10));

You're returning the first number you find. a better approach would be to build the string in the loop then return after that string after the loop. Also no need for num.length if num is an int.
function stringOfEvens(num) {
var stringToReturn = "";
for (let i = 0; i <= num; i++) {
if (i % 2 === 0) {
stringToReturn = stringToReturn + i;
}
}
return stringToReturn;
}

function stringOfEvens(num) {
var str= ' ';
for(var i = 0; i <= num; i = i + 2){
str += i;
}
return str;
}
console.log(stringOfEvens(10))

Just for fun, since it's not particularly efficient:
const stringOfEvens = n => Array(~~(n/2)+1).fill().map((_, i) => 2*i).join("")
or annotated:
const stringOfEvens = n => // arrow function definition, returning:
Array(~~(n / 2) +1) // a sparse array of the required length
.fill() // made unsparse, so .map works
.map((_, i) => 2 * i) // then containing the even numbers
.join("") // and converted to a string
or alternatively (and more efficiently) using Array.from which can call an explicit map function in place:
const stringOfEvens = n => Array.from(Array(~~(n/2)+1), (_, i) => 2*i).join("")

Javascript How to reverse space separated strings

The code below should reverse all the characters in a sentence, but it is unable to do so. This is child's play to me but at this moment it's not compiling. Can anyone figure out the issue?
Let's say:
"Smart geeks are fast coders".
The below code should reverse the above string as follows:
"trams skeeg era tsaf sredoc"
function solution(S){
var result = false;
if(S.length === 1){
result = S;
}
if(S.length > 1 && S.length < 100){
var wordsArray = S.split(" "),
wordsCount = wordsAray.length,
reverseWordsString = '';
for(var i = 0; i< wordsCount; i++){
if(i > 0){
reverseWordsString = reverseWordsString + ' ';
}
reverseWordsString = reverseWordsString + wordsAray[i].split("").reverse().join("");
}
result = reverseWordsString;
}
return result;
}

This should give you the result you're looking for.
function reverseWords(s) {
return s.replace(/[a-z]+/ig, function(w){return w.split('').reverse().join('')});
}

function reverseWords(s) {
var arr = s.split(" ");
s = '';
for(i = 0; i < arr.length; i++) {
s += arr[i].split('').reverse().join('').toLowerCase() + " ";
}
return s;
}
Wouldn't that do the job for you? Basically it just converts the string into an array, by splitting it on space. Then it loops over the array, adds every string reversed to a new string, and then it converts it to lowercase. For faster speed (nothing you would notice), you can just call newStr.toLowerCase() after the loop, so it will do it once instead of every time.

How can I use for loop to count numbers?

I need to count numbers upward and have it print out with a string "then" in between: 5 then 6 then 7 then... like this. I am very confused with using the parameters vs function name when you return. My code is below.. but could someone help with this?
function countUp(start) {
start +=
for(var i = start; i < start + 10; i++) {
console.log(start[i] + "then");
}
return start;
}

I would do something like this:
function countSheep(limit){
for (var i = 1; i < limit; i +=1){
console.log(i + " sheep")
}
}
countSheep(10);
I used "sheep" instead of "then", but you get the idea. Since you just want to produce a side effect (print out a "1 then 2.." to the console, you don;t need to build up a string and then have your function return it.
If you did want to build up a string and then have your function return it though, you could do something like this instead:
function countSheep(limit){
var allMySheep = "";
for (var i = 1; i < limit; i +=1){
allMySheep += (i + " sheep, ")
}
return allMySheep;
}
console.log(countSheep(10));
Note: I started my loops at 1 (var i = 1) because I'm counting sheep, not numbers. You'd probably want to start yours at 0 (var i = 0).

We can use JavaScript join function as well to achieve this
Code
function getCountStr(count) {
var str =[];
for (var i = 1; i <= count; i++) {
str.push(i);
}
console.log(str.join(' then '));
}

There are few issues with your code
function countUp(start) {
start += // <<<<< what's this? It's an incomplete (and useless) statement
for(var i = start; i < start + 10; i++) {
console.log(start[i] + "then");
// ^^^^^^^^ why are doing this? you should only write i
}
return start; // you don't need to return anything
}
A cleaned and working version from your code
function countUp(start) {
for(var i = start; i < start + 10; i++) {
console.log(i + " then ");
}
}
But this code will have an extra 'then' at the end like 1 then 2 then, so here's a code that will handle this
function countUp(start) {
// a temporary array to store your numbers
var tmpArr = [];
for (var i = start; i < start + 10; i++) {
// store the count into the array
tmpArr.push(i);
}
// display the count by putting ' then ' between each number
var stringToDisplay = tmpArr.join(' then ');
console.log(stringToDisplay);
document.write(stringToDisplay);
}
countUp(1);