![\bf (5x-y)^{10}\implies \begin{array}{llll} term&coefficient&value\\ -----&-----&-----\\ 1&+1&(5x)^{10}(-y)^0\\ 2&+10&(5x)^9(-y)^1\\ 3&+45&(5x)^8(-y)^2\\ 4&+120&(5x)^7(-y)^3\\ 5&+210&(5x)^6(-y)^4\\ 6&+252&(5x)^5(-y)^5\\ 7&+210&(5x)^4(-y)^6\\ 8&+120&(5x)^3(-y)^7\\ 9&+45&(5x)^2(-y)^8 \end{array}](/tpl/images/0054/5707/55986.png)
now, how do we get the coefficients?ย well, the first coefficient is 1, any subsequent is " the product of the current terms's coefficient and the exponent of the first element, divided by the exponent of the second element in the next term", now that's a mouthful, but for example,
how did get 210 for the 5th expanded term?ย well is just 120 * 7 / 4
how about 252 of the 6th term?ย ย 210 * 6 / 5.
how about 45 of the 9th one?ย 120 * 3 / 8.
of course, the exponents for each is simple, as you'd already know from the binomial theorem.
so, just expand away the 9th expanded term.