2018-10-10 18:03:03 -04:00
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---
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id: 5900f3a11000cf542c50feb4
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2020-12-16 00:37:30 -07:00
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title: 问题53:组合选择
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2018-10-10 18:03:03 -04:00
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challengeType: 5
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videoUrl: ''
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2021-01-13 03:31:00 +01:00
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dashedName: problem-53-combinatoric-selections
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2018-10-10 18:03:03 -04:00
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---
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2020-12-16 00:37:30 -07:00
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# --description--
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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有十种方法从五种中选择三种,12345:123,124,125,134,135,145,234,235,245和345在组合学中,我们使用符号,5C3 = 10.一般来说,
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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nCr = n!r!(n-r)! ,其中r≤n,n! = n×(n-1)×...×3×2×1和0! = 1。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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直到n = 23,一个值超过一百万:23C10 = 1144066.对于1≤n≤100,nCr的多少,不一定是不同的值大于一百万?
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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# --hints--
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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`combinatoricSelections(1000)`应返回4626。
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2018-10-10 18:03:03 -04:00
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```js
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2020-12-16 00:37:30 -07:00
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assert.strictEqual(combinatoricSelections(1000), 4626);
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2018-10-10 18:03:03 -04:00
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```
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2020-12-16 00:37:30 -07:00
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`combinatoricSelections(10000)`应该返回4431。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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```js
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assert.strictEqual(combinatoricSelections(10000), 4431);
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```
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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`combinatoricSelections(100000)`应返回4255。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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```js
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assert.strictEqual(combinatoricSelections(100000), 4255);
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```
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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`combinatoricSelections(1000000)`应该返回4075。
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2018-10-10 18:03:03 -04:00
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```js
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2020-12-16 00:37:30 -07:00
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assert.strictEqual(combinatoricSelections(1000000), 4075);
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2018-10-10 18:03:03 -04:00
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```
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2020-08-13 17:24:35 +02:00
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2021-01-13 03:31:00 +01:00
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# --seed--
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## --seed-contents--
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```js
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function combinatoricSelections(limit) {
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return 1;
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}
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combinatoricSelections(1000000);
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```
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2020-12-16 00:37:30 -07:00
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# --solutions--
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2021-01-13 03:31:00 +01:00
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```js
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function combinatoricSelections(limit) {
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const factorial = n =>
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Array.apply(null, { length: n })
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.map((_, i) => i + 1)
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.reduce((p, c) => p * c, 1);
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let result = 0;
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const nMax = 100;
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for (let n = 1; n <= nMax; n++) {
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for (let r = 0; r <= n; r++) {
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if (factorial(n) / (factorial(r) * factorial(n - r)) >= limit)
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result++;
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}
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}
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return result;
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}
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```
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