Whats is going on? Python skipping over lines in function -

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When I run the following program in python, takes variables in the function, but completely resume and show anything In addition to the main menu of the program, this qualifying leaves the "if" statement and asks for all the variables, even if the first or second option is selected (which is not required for the third variable). Biteedblu, it should not be an indentation error, I want to show that it was code within Stackevrflo.

Edit: Never. I got it to work. Variables should be identical in function brackets. Oh, is that so! decomposes the forehead 

  Options = 1 option! = 0: print "\ n \ n \ n ************ menu * ************************* ************************************************** ************* print "*" * 28 option = input ( "Please select") # sub-mailing option if option == 1: print "\ n \ n *** **** Calculation of submenu ******* "print" 1. Counting a "print" to 2. Count down from a "Print" 3. Print from various numbers "Print" 4. Print from different numbers * "print" * "* 28 countingmenmen = input (please" Please make a selection: ") x = 0 y = 0 z = 0 q = 0def count (x, y, z, countingSubmenu, q ): x = input ( "Please select your starting number:") Y = input ( "Please select the number of its end:") If countingSubmenu == 1: class Q (x, y + 1, 1): Print q elif countingSubmenu == 2: q (X, Y, -1) in range: print q elif countingSubmenu == 3: z = input ("Please select a increment:") Q (x, y, -z) in the range for print q (x, y + 1, z): and print q: z = input ("please select a increment:"): print q return x , Y, Z, Q If countingSubmenu == 1: counting (CountingSubmenu, X, Y, Z, Q) If countingSubmenu == 2: count (CountingSubmenu, X, Y, Z, Q) if Gnnashubmenu == 3: count ( CountingSubmenu = Count 4: counting (countingSubmenu, X, Y, Z, Q)    
 
 
 your function  count (x, Y, Z, CountingSubmenu, Q)  is defined as, but when you call it, you're argument list  count (Mojhmnu, X, Y, Z , Q) Calculate .

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