freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-53-combinatoric-sel...

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---
id: 5900f3a11000cf542c50feb4
title: 问题53组合选择
challengeType: 5
videoUrl: ''
---
# --description--
有十种方法从五种中选择三种12345123,124,125,134,135,145,234,235,245和345在组合学中我们使用符号5C3 = 10.一般来说,
nCr = nrn-r 其中r≤nn = n×n-1×...×3×2×1和0 = 1。
直到n = 23一个值超过一百万23C10 = 1144066.对于1≤n≤100nCr的多少不一定是不同的值大于一百万
# --hints--
`combinatoricSelections(1000)`应返回4626。
```js
assert.strictEqual(combinatoricSelections(1000), 4626);
```
`combinatoricSelections(10000)`应该返回4431。
```js
assert.strictEqual(combinatoricSelections(10000), 4431);
```
`combinatoricSelections(100000)`应返回4255。
```js
assert.strictEqual(combinatoricSelections(100000), 4255);
```
`combinatoricSelections(1000000)`应该返回4075。
```js
assert.strictEqual(combinatoricSelections(1000000), 4075);
```
# --solutions--