Question

a. In how many different ways you put 20 equal balls, into two numbered boxes?

c. What if you have 20 equal balls, three numbered boxes, and you want to place at least 2 balls in the first box, and no more than 10 balls in the last box?

Answer 1

a. the balls are equal, and the boxes are distinguishalbe so we have

C^20+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=$\sum$¹⁰_k=0$\sum$ ^20−k_j=21

N=$\sum$ ¹⁰_k=0(19−k) ∵$\sum$ ^20−k_j=21=19−k

N=$\sum$ ¹⁰_k=019−$\sum$ ¹⁰_k=0k

N=19($\sum$¹⁰_k=01)−$\sum$¹⁰_k=1k

N=19(11)-((10*11)/2)=209-55=154