---
id: 59f40b17e79dbf1ab720ed7a
title: 部门编号
challengeType: 1
forumTopicId: 302249
dashedName: department-numbers
---

# --description--

There is a highly organized city that has decided to assign a number to each of their departments:

<ul>
  <li>Police department</li>
  <li>Sanitation department</li>
  <li>Fire department</li>
</ul>

Each department can have a number between 1 and 7 (inclusive).

The three department numbers are to be unique (different from each other) and must add up to the number 12.

The Chief of the Police doesn't like odd numbers and wants to have an even number for his department.

# --instructions--

Write a program which outputs all valid combinations as an array.

```js
[2, 3, 7] [2, 4, 6] [2, 6, 4]
[2, 7, 3] [4, 1, 7] [4, 2, 6]
[4, 3, 5] [4, 5, 3] [4, 6, 2]
[4, 7, 1] [6, 1, 5] [6, 2, 4]
[6, 4, 2] [6, 5, 1]
```

# --hints--

`combinations` should be a function.

```js
assert(typeof combinations === 'function');
```

`combinations([1, 2, 3], 6)` should return an Array.

```js
assert(Array.isArray(combinations([1, 2, 3], 6)));
```

`combinations([1, 2, 3, 4, 5, 6, 7], 12)` should return an array of length 14.

```js
assert(combinations(nums, total).length === len);
```

`combinations([1, 2, 3, 4, 5, 6, 7], 12)` should return all valid combinations.

```js
assert.deepEqual(combinations(nums, total), result);
```

# --seed--

## --after-user-code--

```js
const nums = [1, 2, 3, 4, 5, 6, 7];
const total = 12;
const len = 14;
const result = [
  [2, 3, 7],
  [2, 4, 6],
  [2, 6, 4],
  [2, 7, 3],
  [4, 1, 7],
  [4, 2, 6],
  [4, 3, 5],
  [4, 5, 3],
  [4, 6, 2],
  [4, 7, 1],
  [6, 1, 5],
  [6, 2, 4],
  [6, 4, 2],
  [6, 5, 1]
];
```

## --seed-contents--

```js
function combinations(possibleNumbers, total) {

  return true;
}
```

# --solutions--

```js
function combinations(possibleNumbers, total) {
  let firstNumber;
  let secondNumber;
  let thridNumber;
  const allCombinations = [];

  for (let i = 0; i < possibleNumbers.length; i += 1) {
    firstNumber = possibleNumbers[i];

    if (firstNumber % 2 === 0) {
      for (let j = 0; j < possibleNumbers.length; j += 1) {
        secondNumber = possibleNumbers[j];

        if (j !== i && firstNumber + secondNumber <= total) {
          thridNumber = total - firstNumber - secondNumber;

          if (thridNumber !== firstNumber && thridNumber !== secondNumber && possibleNumbers.includes(thridNumber)) {
            allCombinations.push([firstNumber, secondNumber, thridNumber]);
          }
        }
      }
    }
  }
  return allCombinations;
}
```
