## Project Euler 53: Count the values of C(n,r), for 1 ≤ n ≤ 100, that exceed one-million

#### Project Euler 53 Problem Description

Project Euler 53: 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, ^{5}C_{3} = 10.

In general, ^{n}C_{r} = *n*! / *r*!(*n−r*)!, where *r* ≤ *n*, *n*! = *n*×(*n*−1)×…×3×2×1, and 0! = 1.

It is not until *n* = 23, that a value exceeds one-million: ^{23}C_{10} = 1144066.

How many, not necessarily distinct, values of ^{n}C_{r}, for 1 ≤ *n* ≤ 100, are greater than one-million?

#### Analysis

There is an efficient way to solve this using Pascal’s triangle.

1 | ||||||||||||||

1 | 1 | |||||||||||||

1 | 2 | 1 | ||||||||||||

1 | 3 | 3 | 1 | |||||||||||

1 | 4 | 6 | 4 | 1 | ||||||||||

1 | 5 | 10 | 10 | 5 | 1 | |||||||||

1 | 6 | 15 | 20 | 15 | 6 | 1 | ||||||||

1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 |

Each row of this triangle is vertically symmetric, so C(n, r) = C(n, n-r) and any C(n, x) for x from r to (n-r) is greater than C(n, r).

If C(n, r) > 10^{6} then C(n, x) for x from r to (n-r) will also > 10^{6}, therefore, the number of C(n, r) > 10^{6}, is simply (n-r)-r+1 for that row *n*.

#### Project Euler 53 Solution

Runs < 0.001 seconds in Python 2.7.Use this link to get the Project Euler 53 Solution Python 2.7 source.

#### Afterthoughts

- Function
`binomial`

is listed in Common Functions and Routines for Project Euler - Instead of the binomial function you could calculate them on the fly.
- How many, not necessarily distinct, values of
^{n}C_{r}, for 1 ≤*n*≤ 1000, are greater than ten-billion? 490,806

*Project Euler 53 Solution last updated*

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