Counting Problem Consider the equation x+y+z+w=n, where n is a positive integer greater or equal to 4. A positive and integer solution is a set (x,y,z,w) of positive integers that satisfies the equation for a given n. For example, for n=10 one solution to x+y+z+w=10 would be (1,2,3,4), since 1+2+3+4=10. a) Determine the number of positive and integer solutions for n=10. b) Find a general formula that counts the number of whole and integer solutions for x+y+z+w=n. Note: Solutions (2, 2, 2, 4), (2, 2, 4, 2),(2, 4, 2, 2) and (4, 2, 2, 2) for n=10 are each considered different and separate solutions. 
Re: Counting Problem so. its finding the no of partitions of n as 4 integers.. it can be split like partitions of n as sum of two numbers.. and finding the partitions of those 2 numbers as sum of two numbers... 
Re: Counting Problem this can be proved by multinomial theorem... For a+b+c+d = n, the no. of solutions including 0 is need to eliminate those terms in the expansion of that have powers of all a,b,c,d 
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Re: Counting Problem The number of coefficients in the expansion of is (n+m1 n). http://en.wikipedia.org/wiki/Multinomial_theorem So for , it is (n+3 n) ==> 
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((n3)(n2)(n1))/6 I'm completely clueless as to the reasoning behind it, though, and politely ask for a baby explanation lol. 
Re: Counting Problem Hello, Jakarta! Quote:
Place 10 objects in a row, inserting a space between them. [color=beige]. . [/color] Select 3 of the 9 spaces and insert "dividers". Quote:
Following the solution in part (a), place objects in a row (with spaces). Select 3 of the spaces and insert "dividers". 
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Following the solution in part (a), place objects in a row (with spaces). Select 3 of the spaces and insert "dividers". [/quote:3oncju6n] Wow, that is extremely smart and clear. Thanks! 
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