# How do you find the seventh term of (x+y)^12?

Aug 25, 2017

 (12!)/(6!*6!)*x^6y^6.

#### Explanation:

If we expand ${\left(x + y\right)}^{n} ,$ in the descending powers of $x$, the General

${\left(r + 1\right)}^{t h}$ Term, denoted by ${T}_{r + 1} ,$ is given by,

T_(r+1)=""_nC_rx^(n-r)y^r, r=0,1,2,..,n.

In our Problem, $n = 12 , \mathmr{and} , r = 6 ,$ giving,

T_7=""_12C_6x^(12-6)y^6=(12!)/(6!*6!)*x^6y^6.