title: Find the min and max with one term being a sum of an nCr combination of the array and the other term the sum of the remaining items
body: <p>Say you are given an array of signed integers called <code>NUMBERS</code> of length <code>r</code>. Say <code>NUMBERS</code> is equal to</p>
  <p><code>1, 2, 3, 4, 5</code></p>
  <p>where <code>r = 5</code>.</p>
  <p> </p>
  <p>You want to form a sum X of any <code>n</code> elements of the set <code>NUMBERS</code> and multiply this by the sum Y of the remaining <code>m</code> elements for any given <code>n</code>. So the length of the <code>NUMBERS</code> array is <code>(n + m)</code> words in memory.</p>
  <p>This sum is of the form:</p>
  <p><code>(x1 +x2 +x3 +...+xn)*(y1 +y2 +...+ym)</code></p>
  <p> </p>
  <p>For example, with <code>n = 2</code> you could have a possible combination for X be the sum of the arbitrary <code>n = 2</code> elements <code>1</code> and <code>4</code>, and thus the sum Y of the remaining <code>m = 3</code> elements be <code>2 + 3 + 5</code>. You then multiply both sums to get: <code>X * Y = 5 * 10 = 50</code>.</p>
  <p> </p>
  <hr>
  <p>Formally:</p>
  <p>Given <code>(n + m)</code> signed integers, we want to minimize and maximize the result of the algebraic expression of the form:</p>
  <p><code>(x1 +x2 +x3 +...+xn)*(y1 +y2 +...+ym)</code></p>
  <p>for a fixed given <code>n</code>.</p>
  <hr>
  <p> </p>
  <p>This involves finding all <code>nCr</code> combinations of the set <code>NUMBERS</code> where <code>n</code> is given, and then finding the sum X for each combination, then calculating the sum Y of the remaining elements not used for the sum X. Then multiplying both. Using the combinatorics <code>nCr</code> formula you can obtain the number of possible combinations the sum X and the corresponding sum Y, and then brute force every single multiplication to see which will yield the smallest or largest result.</p>
  <p>I am not sure how to proceed, especially since recursion and nested for loops are outside the scope of this first project.</p>