a. the balls are equal, and the boxes are distinguishalbe so we have
C20+2-120=21!/(20!*1!)=21
c. let it k balls in first box,(k≥2), j balls in third box (j≤10), and 20-k-j balls in second box.So
N=10k=0 20−kj=21
N= 10k=0(19−k) ∵ 20−kj=21=19−k
N= 10k=019− 10k=0k
N=19(10k=01)−10k=1k
N=19(11)-((10*11)/2)=209-55=154
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