Answer to Question #148104 in Discrete Mathematics for Promise Omiponle

Question #148104

How many solutions are there to the inequality x1+x2+x3+x4≤15, where x1,x2,x3, and x4 are nonnegative integers?
Hint: Introduce an extra variable x5 and consider x1+x2+x3+x4+x5 = 15.

Expert's answer

"N=C^{15}_{5+15-1}=C^{15}_{19}=\\frac{19!}{15!4!}=3876"

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