An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at most three of them are red is \_\_\_\_\_\_\_\_.
Answer: 490
Solutions · 1
Official solution
At most 3 red means 0, 1, 2, or 3 red marbles drawn (out of 4 total). Equivalently, total ways minus all-4-red:
\[\binom{7}{4} + \binom{5}{1}\binom{7}{3} + \binom{5}{2}\binom{7}{2} + \binom{5}{3}\binom{7}{1} = 35+175+210+70 = 490\]
(Here 7 represents the non-red marbles, 4+3=7.)