(2x-y)6 

Since the exponent is 6 there will be one 
more than 6 terms, or 7 terms.  They are:

C(6,6)(2x)6(-y)0 +
C(6,5)(2x)5(-y)1 + 
C(6,4)(2x)4(-y)2 + 
C(6,3)(2x)3(-y)3 +
C(6,2)(2x)2(-y)4 + 
C(6,1)(2x)1(-y)5 + 
C(6,0)(2x)0(-y)6  

Study the formation of the above 7 terms. 
Notice that each of the 7 terms is of this
form:

C(6,_)(2x)¯(-y)¯

The first two blanks are the same, they start 
with 6 and go down to 0 and the last blank 
starts at 0 and goes up to 6

I will assume you understand how to calculate
the binomial coefficients using this formula:

             n!
C(n,r) = ----------
          r!(n-r)!

where M! = M(M-1)(M-2)···3·2·1

If you don't understand that, post again.
Maybe your book uses nCr
instead but if so, it's the same as C(n,r)

C(6,6)(2x)6(-y)0 =  1(26x6)(-y)0 = 1(64)x6(1) = 64x6
C(6,5)(2x)5(-y)1 =  6(25x5)(-y)1 = 6(32)x5(-y) = -192x5y
C(6,4)(2x)4(-y)2 = 15(24x4)(-y)2 = 15(16)x4y2 = 240x4y2 
C(6,3)(2x)3(-y)3 = 20(23x3)(-y)3 = 20(8)x3(-y3) = -160x3y3
C(6,2)(2x)2(-y)4 = 15(22x2)(-y)4 = 15(4)x2y4 = 60x2y4 
C(6,1)(2x)1(-y)5 =  6(21x1)(-y)5 = 6(2)x1(-y5) = -12xy5
C(6,0)(2x)0(-y)6 =  1(20x0)(-y)6 = 1(1)x0y6 = 1(1)1y6 = y6

Answer:

(2x - y)6 = 
64x6 - 192x5y + 240x4y2 - 160x3y3 + 60x2y4 - 12xy5 + y6 

Edwin