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2) Combinations formula - $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}$

Let us start putting the values of $r$ from $0$ in the formula${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{y^r}} $.$ \Rightarrow {\left( {x + y} \right)^6} = \sum\limits_{r = 0}^6 {^n{C_r}{x^{n - r}}{y^r}} $

We will first put $r = 0$ then $r = 1$ and take it till $r = 6$. Then, we will add all the terms (As formula requires adding the terms.)

$ \Rightarrow {\left( {x + y} \right)^6}{ = ^6}{C_0}{x^{6 - 0}}{y^0}{ + ^6}{C_1}{x^{6 - 1}}{y^1}{ + ^6}{C_2}{x^{6 - 2}}{y^2}{ + ^6}{C_3}{x^{6 - 3}}{y^3}{ + ^6}{C_4}{x^{6 - 4}}{y^4}{ + ^6}{C_5}{x^{6 - 5}}{y^5}{ + ^6}{C_6}{x^{6 - 6}}{y^6}$Now, we know that if $r + p = n$ then,$^n{C_r}{ = ^n}{C_p}$

Using this identity, we can say that $^6{C_0}{ = ^6}{C_6}$, $^6{C_1}{ = ^6}{C_5}$ and$^6{C_2}{ = ^6}{C_4}$.

Thus, we only have to find the value of $^6{C_0}$, $^6{C_1}$, $^6{C_2}$ and $^6{C_3}$.

Let us first find their values individually.

${ \Rightarrow ^6}{C_0} = \dfrac{{6!}}{{0! \times \left( {6 - 0} \right)!}}$$ = \dfrac{{6!}}{{1 \times 6!}} = 1$ …. $(0! = 1)$

Hence, $^6{C_0}{ = ^6}{C_6} = 1$

${ \Rightarrow ^6}{C_1} = \dfrac{{6!}}{{1! \times \left( {6 - 1} \right)!}} = \dfrac{{6!}}{{5!}} = \dfrac{{6 \times 5!}}{{5!}} = 6$

Hence, $^6{C_1}{ = ^6}{C_5} = 6$

${ \Rightarrow ^6}{C_2} = \dfrac{{6!}}{{\left( {6 - 2} \right)! \times 2!}} = \dfrac{{6!}}{{4! \times 2!}} = \dfrac{{6 \times 5 \times 4!}}{{4! \times 2}} = \dfrac{{30}}{2} = 15$

Hence, $^6{C_2}{ = ^6}{C_4} = 15$

${ \Rightarrow ^6}{C_3} = \dfrac{{6!}}{{3! \times \left( {6 - 3} \right)!}} = \dfrac{{6!}}{{3! \times 3!}} = \dfrac{{6 \times 5 \times 4 \times 3!}}{{3! \times 3 \times 2}} = 20$

Now, we will put these values in the expanded expression.

$ \Rightarrow {\left( {x + y} \right)^6} = {x^6} + 6{x^5}y + 15{x^4}{y^2} + 20{x^3}{y^3} + 15{x^2}{y^4} + 6x{y^5} + {y^6}$

Now, since the powers of every term are different, we cannot simplify them further.

You need not memorize the formula. A trick to write the answer is that give the highest power to x and lowest to y. Then, gradually, with the increase in terms, reduce the powers of x by 1 and increase the powers of y by 1. Continue this till the power of x reaches 0 and power of y reaches maximum.