I have a doubt regarding the explanation behind the combinatorial proof of the binomial theorem:
Consider $(x+y)^{4}$. For instance, Why is the coefficient of $x^{2}y^{2}$ the number of ways in which I can select 2 $x$s and 2 $y$s from the four different factors that result from expanding $(x+y)^{4}$?
$(x+ y)^2= (x+ y)(x+ y)= x(x+ y)+ y(x+ y)= x^2+ **xy**+ **yx**+ y^2= x^2+ **2**xy+ y^2$. The "2" is from adding "xy" and "yx".
$(x+ y)^3= (x+ y)(x^2+ 2x+ y^2)= x(x^2+ 2xy+ y^2)+ y(x^2+ 2xy+ y^2)= x^3+ **2x^2y**+ **xy^2**+ **x^2y**+ **2xy^2**+ y^3= x^3+ **3**x^2y+ **3**xy^2+ y^3$ The first "3" is from adding "$2x^2y" and "$x^2y$" and the second is from adding "$xy^2$" and "$2xy^2$".
$(x+ y)^4= (x+ y)(x^3+ 3x^2y+ 3xy^2+ y^3)= x(x^3+ 3x^2y+ 3xy^2+ y^3)+ y(x^3+ 3x^2y+ 3xy^2+ y^3)= x^4+ **3x^3y**+ **3x^2y^2**+ **xy^3**+ **x^3y**+ **3x^2y^2**+ **3xy^3**+ y^4= x^4+ **4**x^3y+ **6**x^2y^2+ **4**x^3y+ y^4$ The first "4" is from adding "$3x^3y$" and "$x^3y$" and the second is from adding "$xy^3$" and "$3xy^3$". The "6" is from adding "$3x^2y^2$" and "$3x^2y^2$".