#53 - Combinatoric selections

17 Jun 2016 in Project Euler on 5%

There are exactly ten ways of selecting three from five, 12345: $$123,\,\,124,\,\,125,\,\,134,\,\,135,\,\,145,\,\,234,\,\,235,\,\,245,\,\,\text{and }345$$

In combinatorics, we use the notation $_5 C_3=10$.

In general $_nC_r=\frac{n!}{r!(n-r)!}$, where $r\leq n, n!=n\times(n-1)\times\dots\times 3\times 2\times 1$, and $0!=1$.

How many, not necessarily distinct, values of $_nC_r$ for $1\leq n\leq 100$, are greater than one-million?

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Notice that the formula for $_nC_r$ has three factorials. Recomputing each of these factorials each time will be time-consuming and inefficient. Instead, we can find $n!$ for all $0\leq n\leq 100$, and grab the values we need for each computation. Python’s math package has a quick factorial function we can use:

# file: "problem053.py"
factorials = [math.factorial(i) for i in range(0, 101)]
count = 0
for n in range(23, 101):
    for r in range(n + 1):
        if factorials[n] / (factorials[r] * factorials[n - r]) > 1000000:
            count += 1
print(count)

Running this short loop results in,

4075
0.0038949089640585117 seconds.

Thus, there are 4075 values that are greater than one million.

Bonus

We can actually solve this problem without computing a single factorial! We can use Pascal’s triangle to find $_nC_r$, as the rows of the triangle are the values of $_nC_r$. Pascal’s triangle is constructed in the following manner:

1. The first row is $1$.
2. The second row is $1\quad 1$.
3. Each row after the second is formed by placing two ones at the ends, while the middle values are calculated by adding the two numbers above. For example, the first 5 rows are shown below.

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

If the rows are numbered starting from 0, and the elements in the row are also numbered starting from 0, the $r^\text{th}$ number in the $n^\text{th}$ row is exactly $_nC_r$. Using the generative nature of Pascal’s triangle, we can continuously generate each row until the 100th row, and count the numbers which are greater than one-million.

currRow = [1, 1]
count = 0
for n in range(2, 101):
    nextRow = [0] * (n + 1)
    nextRow[0] = 1
    nextRow[-1] = 1
    # Make the next row
    for i in range(len(currRow) - 1):
        nextRow[i+1] = currRow[i] + currRow[i+1]
    # Count how many are bigger than 1 million
    count += len([x for x in nextRow if x > 1000000])
    currRow = nextRow
print(count)

We have to start from the beginning in order to see what the 23rd row is. Regardless, the output of is the same, although slightly faster.

4075
0.0015237512804923096 seconds.