Data commit

Task/Evaluate-binomial-coefficients/MATLAB/evaluate-binomial-coefficients-1.m

@@ -0,0 +1,3 @@
+>> nchoosek(5,3)
+ans =
+    10

Task/Evaluate-binomial-coefficients/MATLAB/evaluate-binomial-coefficients-2.m

@@ -0,0 +1,4 @@
+function r = binomcoeff1(n,k)
+    r = diag(rot90(pascal(n+1))); % vector of all binomial coefficients for order n
+    r = r(k);
+end;

Task/Evaluate-binomial-coefficients/MATLAB/evaluate-binomial-coefficients-3.m

@@ -0,0 +1,3 @@
+function r = binomcoeff2(n,k)
+   prod((n-k+1:n)./(1:k))
+end;

Task/Evaluate-binomial-coefficients/MATLAB/evaluate-binomial-coefficients-4.m

@@ -0,0 +1,4 @@
+function r = binomcoeff3(n,k)
+   m = pascal(max(n-k,k)+1);
+   r = m(n-k+1,k+1);
+end;

Task/Evaluate-binomial-coefficients/MATLAB/evaluate-binomial-coefficients-5.m

@@ -0,0 +1,21 @@
+function coefficients = binomialCoeff(n,k)
+
+    coefficients = zeros(numel(n),numel(k)); %Preallocate memory
+
+    columns = (1:numel(k)); %Preallocate row and column counters
+    rows = (1:numel(n));
+
+    %Iterate over every row and column. The rows represent the n number,
+    %and the columns represent the k number. If n is ever greater than k,
+    %the nchoosek function will throw an error. So, we test to make sure
+    %it isn't, if it is then we leave that entry in the coefficients matrix
+    %zero. Which makes sense combinatorically.
+    for row = rows
+        for col = columns
+            if k(col) <= n(row)
+                coefficients(row,col) = nchoosek(n(row),k(col));
+            end
+        end
+    end
+
+end %binomialCoeff

Task/Evaluate-binomial-coefficients/MATLAB/evaluate-binomial-coefficients-6.m

@@ -0,0 +1,40 @@
+>> binomialCoeff((0:5),(0:5))
+
+ans =
+
+     1     0     0     0     0     0
+     1     1     0     0     0     0
+     1     2     1     0     0     0
+     1     3     3     1     0     0
+     1     4     6     4     1     0
+     1     5    10    10     5     1
+
+>> binomialCoeff([1 0 3 2],(0:3))
+
+ans =
+
+     1     1     0     0
+     1     0     0     0
+     1     3     3     1
+     1     2     1     0
+
+>> binomialCoeff(3,(0:3))
+
+ans =
+
+     1     3     3     1
+
+>> binomialCoeff((0:3),2)
+
+ans =
+
+     0
+     0
+     1
+     3
+
+>> binomialCoeff(5,3)
+
+ans =
+
+    10