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All the colors in (Posted on 2010-04-26) Difficulty: 3 of 5
Seven sportsmen (named A,B,C,D,E,F and G) buy 7 hats choosing randomly out of 4 available colors (say G,R,W and Y).
What is the probability that all the colors were chosen?

  Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)
Solution: (Hide)
7 can be expressed as a sum of 4 positive integers in 3 ways:
a. 7=1+1+1+4
b. 7=1+1+2+3
c. 7=1+2+2+2
We shall separately calculate the number of different colors allocation for each of the above cases.

Case a: We can choose the 4 people getting a W-colored hat in 7!/(3!*4!) ways distributing one hat of each non-W color in 3! ways to the remaining sportsmen.
So there are 7!/(3!*4!)*3! ways if we start with a W color and 4 times as much if we it is any color that 4 people are getting.
Therefore the quantity of combinations
for case a is 7!/(3!*4!)*3! *4=840.

Case b: Let's choose what couple gets W and Y hats . First 3 are out of 7 get W, then 2 are chosen out of the 4 remaining to get color Y. There are 7!/(3!*4!)*4! /(2!*2!) ways to achieve this with W-Y selected colors and 12(=4*3) times as much if all possible couples are considered. We have to give the 2 remaining colors to 2 remaining sportsmen, just doubling the number of permissible selections.
Therefore the quantity of combinations for case b
is 7!/(3!*4!)*4!/(2!*2!)*12*2=5040.

Case c: Let's choose who gets 1 W hat . We have 7 choices to do so and 6!/(2!*4!)*4!/(2!*2!) *3! for the rest. We have 4 times as much if we choose any color as a starter not just W.
Therefore the quantity of combinations for case c is
7*6!/(2!*4!)*4!/(2!*2!)*4 =2520.

All three cases being exclusive , we just sum-up 840+5040+2520= 8400.

Answer: 8400

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: All the colors solutionAdy TZIDON2010-08-21 11:31:09
All the colors solutionDan Rosen2010-06-16 12:41:08
SolutionPraneeth2010-05-10 09:19:54
SolutionSolution (using Excel)hoodat2010-05-03 02:03:31
re(2): SolutionJohn zadeh2010-04-28 16:14:09
re: SolutionCharlie2010-04-28 11:12:54
SolutionJohn zadeh2010-04-27 19:43:38
SolutionJohn zadeh2010-04-27 19:42:29
SolutionJohn zadeh2010-04-27 19:41:58
SolutionJohn zadeh2010-04-27 19:40:42
SolutionJohn zadeh2010-04-27 19:40:29
SolutionJohn zadeh2010-04-27 19:40:11
SolutionJohn zadeh2010-04-27 19:39:54
Hints/Tipsre: SolutionAdy TZIDON2010-04-27 19:38:36
SolutionJohn zadeh2010-04-27 19:38:16
SolutionJohn zadeh2010-04-27 19:37:22
SolutionJohn zadeh2010-04-27 19:30:43
SolutionJohn zadeh2010-04-27 19:30:10
SolutionsolutionCharlie2010-04-26 13:50:11
Some Thoughtsapproximation (spoiler)Steve Herman2010-04-26 13:43:43
mad hattters?ed bottemiller2010-04-26 13:28:36
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