How to pass model to partialview at run time? - asp.net-mvc

Question

ViewModel
public class ModelTypeViewModel
{
public virtual CheckRadioButton CRB { get; set; }
}
Controller
public class M1Controller : Controller
{
public CarContext db = new CarContext();
private CheckRadioButton get()
{
CheckRadioButton c = new CheckRadioButton();
c.BrandName = "abc";
c.type = "xyz";
return c;
}
public ActionResult Hello ()
{
CheckRadioButton s = get();
ModelTypeViewModel mm = new ModelTypeViewModel(s);
return View(mm);
}
View:(Hello)
#model Car.Models.ModelTypeViewModel
#Html.Partial("_Display", Model.CRB)
Partial View(_Display)
<h1> Hello </h1>
How can I pass diff model each time to to partial view?
It gives an error
"An exception of type 'System.Web.HttpParseException' occurred in System.Web.WebPages.Razor.dll but was not handled in user code"
It gives the same error even if I pass only 'Model"
I am confused

Answer 1

Put each button inside Ajax.BeginForm
#using (Ajax.BeginForm("BuyItem", "MsmqTest"}, new AjaxOptions { UpdateTargetId = "msmqpartial" }))
{ <button type="submit">Buy</button>}
#using (Ajax.BeginForm("BuyItem", "MsmqTest" }, new AjaxOptions { UpdateTargetId = "msmqpartial" }))
{
<button type="submit">Sell</button>
}
Where "updateTargetId" is the div id to append content
public ActionResult BuyItem()
{
if(//some condition goes here)
return PartialView("Partial1",data);
if(//some condition goes here)
return PartialView("Partial2",data);
}