2.3 KiB
2.3 KiB
id | challengeType | title | forumTopicId |
---|---|---|---|
5900f3a11000cf542c50feb4 | 5 | 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) {
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;
}