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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, 5C3 = 10.
In general, nCr = 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: 23C10 = 1144066.
How many, not necessarily distinct, values of nCr, 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) > 106 then C(n, x) for x from r to (n-r) will also > 106, therefore, the number of C(n, r) > 106, is simply (n-r)-r+1 for that row n.
Project Euler 53 Solution
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Use this link to get the Project Euler 53 Solution Python 2.7 source.
Afterthoughts
- Function
binomialis 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 nCr, for 1 ≤ n ≤ 1000, are greater than ten-billion? 490,806
Discussion
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