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Vedic Math Techniques informal methods such as chunking, grids, number lines and visual method :)
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Well I really don't get the question but I'll answer it according to my under standing

let say we got this binomial equation in terms of x and y 


first the number of terms is the expansion is n+1
the first term is x^n and the last term is y^n
the exponent of x descends linearly from n to 0
the exponent of y descends linearly from 0 to n
the sum of the exponents of a and b in any of the terms is equal to n,
the coefficient of the second term and the second to the last term is n,


(x+y)^n=(x^ny^0+n x^{n-1} y^1+ \frac{n}{2}  x^{n-2} y^2+...+ \frac{n}{n-1}x^1 y^{n-1}  + \frac{n}{n} x^{0}   y^{n} )



in this example n = 3

the number of therms is 3 + 1 = 4

so there are 4 terms in this example

 (x+y)^{3} = ( x^{3}y^0 +3 x^{3-1}  y^{1} +3x y^{3-1} + \frac{3}{3}y^3)

 (x+y)^{3} =  x^{3}  y^{0} +3 x^{2}  y^{1} +3 x^{1}  y^{2} +  x^{0}  y^{3}

see the pattern in which the exponent of x is decreasing from 3 to 0
and the pattern in which the exponent of y is increasing from  0 to 3

this is the pattern that I knew in multiplying binomial algebraic expression

don't know if this is the answer you are looking for. 

hit thank you if this helps you..
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