How Many Solutions(x, Y, Z,w) Are There To Theequa


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How many solutions(x, y, z,w) are there to theequation x+y+z+w= 2009 wherex, y, z, w are nonnegative integers.


Answer No.1


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Number of non-negative solutions forx1+x2+x3+....+xr=n isn+r-1Cr-1 given case we have r=4 and n=2009 hence the answer is 2009+4-1C4-1 =2012C3 =1355454220

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