freeCodeCamp/curriculum/challenges/spanish/08-coding-interview-prep/project-euler/problem-53-combinatoric-selections.spanish.md

id, localeTitle, challengeType, title

id	localeTitle	challengeType	title
5	5900f3a11000cf542c50feb4	5	Problem 53: Combinatoric selections

Description

Hay exactamente diez formas de seleccionar tres de cinco, 12345: 123, 124, 125, 134, 135, 145, 234, 235, 245 y 345 En combinatoria, usamos la notación, 5C3 = 10. In general,

nCr = n! r! (n − r)! , donde r ≤ n, n! = n × (n − 1) × ... × 3 × 2 × 1, y 0! = 1.

No es hasta n = 23, que un valor excede de un millón: 23C10 = 1144066. ¿Cuántos, no necesariamente distintos, valores de nCr, para 1 ≤ n ≤ 100, son mayores que un millón? ?

Instructions

Tests

tests:
  - text: <code>combinatoricSelections(1000)</code> deben devolver 4626.
    testString: 'assert.strictEqual(combinatoricSelections(1000), 4626, "<code>combinatoricSelections(1000)</code> should return 4626.");'
  - text: <code>combinatoricSelections(10000)</code> deben devolver 4431.
    testString: 'assert.strictEqual(combinatoricSelections(10000), 4431, "<code>combinatoricSelections(10000)</code> should return 4431.");'
  - text: <code>combinatoricSelections(100000)</code> deben devolver 4255.
    testString: 'assert.strictEqual(combinatoricSelections(100000), 4255, "<code>combinatoricSelections(100000)</code> should return 4255.");'
  - text: <code>combinatoricSelections(1000000)</code> deben devolver 4075.
    testString: 'assert.strictEqual(combinatoricSelections(1000000), 4075, "<code>combinatoricSelections(1000000)</code> should return 4075.");'

Challenge Seed

function combinatoricSelections(limit) {
  // Good luck!
  return 1;
}

combinatoricSelections(1000000);

Solution

function combinatoricSelections(limit) {
    const factorial = n =>
        Array.apply(null, { length: n })
            .map((_, i) => i + 1)
            .reduce((p, c) => p * c, 1);

    let result = 0;
    const nMax = 100;

    for (let n = 1; n <= nMax; n++) {
        for (let r = 0; r <= n; r++) {
            if (factorial(n) / (factorial(r) * factorial(n - r)) >= limit)
                result++;
        }
    }

    return result;
}