How to choose random N item from an array in Javascript without repetition?

Question

I have an array containing some items, let's say numbers:

[23,4,67,8,29,46,7,2,98]

How it is possibile to choose N different random number (no more, no less) from the array?

Behaviour expected: if N=3

[4,23,98] accepted
[7,67,29] accepted
[7,67,7]  not accepted

EDIT: Finally I've found a pretty good solution. The thread as been closed so I'm publishing my solution here just for people passing by. I had an array long M and N numbers to choose so I made an array containing M numbers (indexesArray) applied the Fisher Yates algorithm to shuffle number corresponding this way to indexes of my initial array and took the first M items from that. Thanks to @ArneHugo for pointing the right path to follow.

var arrayContainingNumbersToChoose=[23,4,67,8,29,46,7,2,98];
var indexesArray=[];
var N=5;
var resultArray=[];

var M=arrayContainingNumbersToChoose.length;

for (let i = 0; i < M; i++) {
   indexesArray.push(i);
} // result like [0, 1, 2, 3, 4, 5, 6, 7, 8]

indexesArray=fisherYatesShuffle(indexesArray); // result like [2, 4, 7, 6, 0, 1, 3, 8, 5]

for (let i = 0; i < N; i++) {
resultArray.push(arrayContainingNumbersToChoose[indexesArray[i]]);
}

console.log(resultArray);

function fisherYatesShuffle(a) {
      var j, x, i;
      for (i = a.length - 1; i > 0; i--) {
          j = Math.floor(Math.random() * (i + 1));
          x = a[i];
          a[i] = a[j];
          a[j] = x;
      }
      return a;
  }

Answer 1

Try the following:

var arr = [23,4,67,8,29,46,7,2,98];
var result = [];
var map = {};
var n = 3;
var i = arr.length;
var counter = arr.length+n;
while(result.length != n){
      num = arr[Math.floor(Math.random() * i)];
      if(!map[num]){
        map[num] = i;
        result.push(num);
      }
      if(counter <= 0) // stop finding random numbers if counter is zero.
        break;
      counter--;
      i--;
      if(i  == 0)
        i = arr.length;
}
console.log(result);