diff --git a/Computer Oriented Statistical Methods/Gauss_Elimination.c b/Computer Oriented Statistical Methods/Gauss_Elimination.c
new file mode 100644
index 00000000..e35fa7ba
--- /dev/null
+++ b/Computer Oriented Statistical Methods/Gauss_Elimination.c	
@@ -0,0 +1,100 @@
+#include<stdio.h>
+#include<conio.h>
+void display(float a[20][20],int n)
+{
+	  int i,j;
+	  for(i=0;i<n;i++)
+	  {
+		    for(j=0;j<=n;j++)
+		    {
+			      printf("%.2f \t",a[i][j]);
+		    }
+		    printf("\n");
+	  }
+}
+float interchange(float m[20][20],int i,int n)
+{
+	  float tmp[20][20];
+	  float max=fabs(m[i][i]);
+	  int j,k=i;
+
+	  for(j=i;j<n;j++)
+	  {
+		    if(max<fabs(m[j][i]))
+		    {
+			      max=fabs(m[j][i]);
+			      k=j;
+		    }
+	  }
+	  for(j=0;j<=n;j++)
+	  {
+		    tmp[i][j]=m[i][j];
+		    m[i][j]=m[k][j];
+		    m[k][j]=tmp[i][j];
+	  }
+	  return m[20][20];
+}
+float eliminate(float m[20][20],int i,int n)
+{
+	float tmp;
+	int k=1,l,j;
+	for(j=i;j<n-1;j++)
+	{
+		  tmp=-((m[i+k][i])/(m[i][i]));
+		  for(l=0;l<=n;l++)
+		  {
+			    m[i+k][l]=(m[i+k][l])+(m[i][l]*tmp);
+		  }
+		  k++;
+	}
+	return m[20][20];
+}
+void main()
+{
+	  int i,j,n,k=0,l;
+	  float m[20][20],mul,tmp[20][20],val,ans[20];
+	  clrscr();
+
+	  printf("Total No.of Equations : ");
+	  scanf("%d",&n);
+
+	  printf("\n");
+	  for(i=0;i<n;i++)
+	  {
+
+		    printf("Enter Co-efficient Of Equations %d & Total --->>>\n",i+1);
+		    for(j=0;j<=n;j++)
+		    {
+			      printf("r%d%d : ",i,j);
+			      scanf("%f",&m[i][j]);
+		    }
+		    printf("\n");
+	  }
+	  printf(":::::::::::: Current Matrix ::::::::::::\n\n");
+	  display(m,n);
+
+	 for(i=0;i<n-1;i++)
+	 {
+		    printf("\n------->>>>>>>>>>>>>>>>>>>>>>>>-------- %d\n",i+1);
+		    m[20][20]=interchange(m,i,n);
+		    display(m,n);
+		    printf("\n_______________________________________\n");
+		    m[20][20]=eliminate(m,i,n);
+		    display(m,n);
+	  }
+	  printf("\n\n Values are : \n");
+	 for(i=n-1;i>=0;i--)
+	  {
+		    l=n-1;
+		    mul=0;
+		    for(j=0;j<k;j++)
+		    {
+			    mul=mul+m[i][l]*ans[l];
+			    l--;
+		    }
+		    k++;
+		    ans[i]=(m[i][n]-mul)/m[i][i];
+		    printf("X%d = %.2f\n",i+1,ans[i]);
+	  }
+getch();
+}