# JavaScript - Count all the numbers with unique digits in a range

## JavaScript Math: Exercise-109 with Solution

Write a JavaScript program that accepts a number (n) and counts all numbers with unique digits of length p within a specified range.

Range: 0 <= p < 10^{n}

Example:

When n = 1, numbers with unique digits (10) between 0 and 9 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

When n = 2, numbers with unique digits (91) between 0 and 100 are 0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15 …..99 except 11, 22, 33, 44, 55, 66, 77, 88 and 99.

**Test Data:**

(1) -> 10

(2) -> 91

**Sample Solution:**

**HTML Code:**

```
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>JavaScript program to Count all numbers with unique digits in a range</title>
</head>
<body>
</body>
</html>
```

**JavaScript Code:**

```
function test(n) {
if(n===0){
return 1;
}
var result = 10;
var temp =9;
for(var i=1; i<n; i++){
temp *= 10-i
result+=temp
}
return result;
}
n = 1
console.log("Range: "+n+" to 10")
console.log("Numbers with unique digits in the said range: "+test(n));
n = 2
console.log("Range: "+n+" to 10")
console.log("Numbers with unique digits in the said range: "+test(n));
```

Sample Output:

Range: 1 to 10 Numbers with unique digits in the said range: 10 Range: 2 to 10 Numbers with unique digits in the said range: 91

**Flowchart: **

**Live Demo: **

See the Pen javascript-math-exercise-109 by w3resource (@w3resource) on CodePen.

**Improve this sample solution and post your code through Disqus**

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