Problem
I. An urn contains: 4 blue balls, 5 red balls, 2 green balls and 3 pink balls. 3 balls are drawn successively and without replacement. 1) How many possible outcomes can be formed? 2) How many possible outcomes can be formed such that: a) The first ball is pink, the second ball is red and the third ball is blue. b) The three balls are red. c) The three balls are green. d) Two balls are red and one ball is pink. e) One ball is pink, one is blue and one is red. f) At most one ball is red. g) At least one ball is blue. h) No red balls. i) Three balls of the same color.
Solution
Step 1 ：1) \( \frac{14!}{11! \cdot 3!} = {^{14} C _3} = 364 \)
Step 2 ：2a) \( 3 \cdot 5 \cdot 4 = 60\)
Step 3 ：2b) \( {^5 C _3} = 10\)
Step 4 ：2c) \( {^2 C _3} = 0\)
Step 5 ：2d) \( {^5 C _2} \cdot {^3 C _1} = 60\)
Step 6 ：2e) \( 3 \cdot 4 \cdot 5 =180\)
Step 7 ：2f) \( 364 - {^{5} C _2} \cdot {^9 C _1} + 5 = 214\)
Step 8 ：2g) \( 364 - {^5 C _3} \cdot {^9 C _0} = 354\)
Step 9 ：2h) \( {^9 C _3} = 84\)
Step 10 ：2i) \( 10 + 0 + {^4 C _3} = 14\)
