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Project Euler Solutions

Project Euler 53 Solution

Project Euler 53 Solution

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 rn, 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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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  nCr, for 1 ≤ n ≤ 1000, are greater than ten-billion? 490,806
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