编写此jquery代码的更有效方法

有没有更好、更有效的方法来编写这段代码?这是一个临时的下拉菜单,允许用户对多人进行rsvp。对不起,有点乱,但我想我做的很清楚。如果没有,我在我的电脑,将迅速作出反应,更多的信息需要。

 //There's got to be a better way to do this
 $('#guest_num_1').click( function() {
 $('#num_guests a#quant_guests').html("1")
 $('.guest_name_2, .guest_name_3, .guest_name_4, .guest_name_5, .guest_name_6 ').hide()
 });

 $('#guest_num_2').click( function() {
 $('#num_guests a#quant_guests').html("2")
 $('.guest_name_2').fadeIn()
 $('.guest_name_3, .guest_name_4, .guest_name_5, .guest_name_6').hide()
 });

 $('#guest_num_3').click( function() {
 $('#num_guests a#quant_guests').html("3")
 $('.guest_name_2, .guest_name_3').fadeIn()
 $('.guest_name_4, .guest_name_5, .guest_name_6').hide()
 });

 $('#guest_num_4').click( function() {
 $('#num_guests a#quant_guests').html("4")
 $('.guest_name_2, .guest_name_3, .guest_name_4').fadeIn()
 $('.guest_name_5, .guest_name_6').hide()
 });

 $('#guest_num_5').click( function() {
 $('#num_guests a#quant_guests').html("5")
 $('.guest_name_2, .guest_name_3, .guest_name_4, .guest_name_5').fadeIn()
 $('.guest_name_6').hide()
 });

 $('#guest_num_6').click( function() {
 $('#num_guests a#quant_guests').html("6")
 $('.guest_name_2, .guest_name_3, .guest_name_4, .guest_name_5, .guest_name_6').fadeIn()
 });