freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-53-combinatoric-selections.english.md

id, challengeType, isHidden, title, forumTopicId

id	challengeType	isHidden	title	forumTopicId
5900f3a11000cf542c50feb4	5	false	Problem 53: Combinatoric selections	302164

Description

There are exactly ten ways of selecting three from five, 12345:

123, 124, 125, 134, 135, 145, 234, 235, 245, and 345

In combinatorics, we use the notation, \displaystyle \binom 5 3 = 10

In general, \displaystyle \binom n r = \dfrac{n!}{r!(n-r)!}, where r \le n, n! = n \times (n-1) \times ... \times 3 \times 2 \times 1, and 0! = 1.

It is not until n = 23, that a value exceeds one-million: \displaystyle \binom {23} {10} = 1144066.

How many, not necessarily distinct, values of  \displaystyle \binom n r for 1 \le n \le 100, are greater than one-million?

Instructions

Tests

tests:
  - text: <code>combinatoricSelections(1000)</code> should return a number.
    testString: assert(typeof combinatoricSelections(1000) === 'number');
  - text: <code>combinatoricSelections(1000)</code> should return 4626.
    testString: assert.strictEqual(combinatoricSelections(1000), 4626);
  - text: <code>combinatoricSelections(10000)</code> should return 4431.
    testString: assert.strictEqual(combinatoricSelections(10000), 4431);
  - text: <code>combinatoricSelections(100000)</code> should return 4255.
    testString: assert.strictEqual(combinatoricSelections(100000), 4255);
  - text: <code>combinatoricSelections(1000000)</code> should return 4075.
    testString: assert.strictEqual(combinatoricSelections(1000000), 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;
}