Question wrt Binomial theorem

Question

This question may sound too stupid. But I am quite confused by the Binomial theorem

As per my understanding, let us consider

$(x+y)^4$ without the coefficients.

$(x+y)^4 = x^4 + x^3y + x^2y^2 + xy^3 + y^4$

Now if we take the term $x^2y^2$ as per the binomial theorem the coefficient would be 6. because ${4 \choose 2}$ gives the answer to be 6. But why do we take only $y$ term into consideration during coefficient calculation but now x term.

I feel it should be $4 \choose 2$$4 \choose 2$ both multiplied because we need to select two $x$ and two $y$. In that case the coefficient would be 36. But according to binomial theorem it is 6. Where I am going wrong?

kindly explain me

Answer 1

Note that$$(x+y)^4=(x+y)\times(x+y)\times(x+y)\times(x+y).$$The coefficient of $xy$ is the number of $x^2y^2$'s that we get when we expand this product. And there are as many ways of getting a $x^2y^2$ as there are ways of picking two of the factors and using its $x$ and using the $y$ from the remaining two factors. And in how many ways can we do that? In $\binom42=6$ ways.

Answer 2

The selection is symmetric in $x$ and in $y$. In order to count the number of ways to get $x^2y^2$ in \begin{align*} (x+y)^4=(x+y)(x+y)(x+y)(x+y) \end{align*} we select from each of the four binomials $x+y$ either $x$ or $y$ in total two $x$'s and two $y$'s. We have \begin{align*} &x\,x\,\color{blue}{y}\,\color{blue}{y}\qquad\qquad \color{blue}{y}\,\color{blue}{y}\,x\,x\\ &x\,\color{blue}{y}\,x\,\color{blue}{y}\qquad\qquad \color{blue}{y}\,x\,\color{blue}{y}\,x\\ &x\,\color{blue}{y}\,\color{blue}{y}\,x\qquad\qquad \color{blue}{y}\,x\,x\,\color{blue}{y} \end{align*}

giving a total of $6$ possibilities.

In fact this symmetry is also given by the definition of the binomial coefficient in terms of factorials \begin{align*} 6=\binom{4}{2}=\frac{4!}{\color{blue}{2!2!}}=\binom{4}{\color{blue}{2,2}} \end{align*} and also explicitly by the multinomial coefficient notation at the right-hand side.