Question 1158319
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Picture the expression as 9 factors of (2x+3y) being multiplied together.<br>
The first term in the product will use the x term from all 9 factors;
The second term will use the x term 8 times and the y term 1 time;
...
The fifth term will use the x term 5 times and the y term 4 times.<br>
The first part in the calculation of the fifth term is determining the number of different ways you can choose 5 of the 9 factors to be the ones in which the x term is chosen.  That number of ways is "9 choose 5":<br>
{{{C(9,5) = 9!/((5!)(9-5)!) = (9*8*7*6)/(4*3*2*1) = 84}}}<br>
In each of those 84 cases, the (2x) factor is used 5 times and the (3y) factor is used 4 times.<br>
So the fifth term in the expansion is<br>
{{{(84)((2x)^5)((3y)^4)}}} = {{{((84)(2^5)(3^4))x^5y^4}}}<br>
Use a calculator if you need to multiply all that out.<br>