All the colors in (Posted on 2010-04-26)

	Submitted by Ady TZIDON
Rating: 5.0000 (1 votes)
Solution:	(Hide)
Solution: 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

	Subject	Author	Date
	Puzzle Answer	K Sengupta	2023-10-04 11:00:09
	re: All the colors solution	Ady TZIDON	2010-08-21 11:31:09
	All the colors solution	Dan Rosen	2010-06-16 12:41:08
	Solution	Praneeth	2010-05-10 09:19:54
	Solution (using Excel)	hoodat	2010-05-03 02:03:31
	re(2): Solution	John zadeh	2010-04-28 16:14:09
	re: Solution	Charlie	2010-04-28 11:12:54
	Solution	John zadeh	2010-04-27 19:43:38
	Solution	John zadeh	2010-04-27 19:42:29
	Solution	John zadeh	2010-04-27 19:41:58
	Solution	John zadeh	2010-04-27 19:40:42
	Solution	John zadeh	2010-04-27 19:40:29
	Solution	John zadeh	2010-04-27 19:40:11
	Solution	John zadeh	2010-04-27 19:39:54
	re: Solution	Ady TZIDON	2010-04-27 19:38:36
	Solution	John zadeh	2010-04-27 19:38:16
	Solution	John zadeh	2010-04-27 19:37:22
	Solution	John zadeh	2010-04-27 19:30:43
	Solution	John zadeh	2010-04-27 19:30:10
	solution	Charlie	2010-04-26 13:50:11
	approximation (spoiler)	Steve Herman	2010-04-26 13:43:43
	mad hattters?	ed bottemiller	2010-04-26 13:28:36