
Paul F. answered 08/08/17
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Sure there is.
(2x+y)5 = C (5,0) (2x)5 y0 + C (5,1) (2x)4 y1 + C (5,2) (2x)3 y2 + C (5,3) (2x)2 y3 + ........ C (5,5) (2x)0 y5
The 4th term variables are x2 y3 and the coefficient = 4 [ C(5,3) ] = 40
Karen L.
08/08/17