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dan815
 one year ago
there are a 103 choose 3 positive integer solutions to a+b+c+d = 100.
find the number of positive integer solutions to
a+b+2c+2d=100, a,b,c,d>=0
dan815
 one year ago
there are a 103 choose 3 positive integer solutions to a+b+c+d = 100. find the number of positive integer solutions to a+b+2c+2d=100, a,b,c,d>=0

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Question about your question, so I can choose any 4 values for a b c and d?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you sure there are 103 choose 3 positive integer solutions to a+b+c+d = 100 ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ya with 0 included i think

dan815
 one year ago
Best ResponseYou've already chosen the best response.0do you find something wrong with that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I believe it should be \(\dbinom{99}3\). https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)#Statements_of_theorems

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i can see why 99 choose 3 is right, for some reason i also cant find why this way of thinking is wrong even if its the wrong answer like for 3s and 2 bars 1,1,1,, there are 5 spots and i have to pick 2 spots as bars so 5 Choose 2

dan815
 one year ago
Best ResponseYou've already chosen the best response.0can you explain what is wrong with that process

dan815
 one year ago
Best ResponseYou've already chosen the best response.0hmm wait are u sure 99 choose 3 is right htough

dan815
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles 99choose3 is right if 0 is not allowed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There are a 99 choose 3 'positive integer solutions' to a+b+c+d = 100. This is because you say 'positive integer solutions, so we exclude zero. If we include zero then it is 103 choose 3= 176847 , I checked with python and both are correct Now as far as your question above, it's better if we say find the number of nonnegative integer solutions to a+b+2c+2d=100, a,b,c,d>=0 Or you can say, find the number of integer solutions to a + b+ 2c + 2d = 100 a>=0, b>=0, c>=0, d>=0 If you say find the number of positive integer solutions, you really mean to solve a + b+ 2c + 2d = 100 a >0 , b > 0 , c > 0 , d > 0 Using python I got 45526 code: count = 0 for a in range(0,101): for b in range(0,101): for c in range(0,51): for d in range (0,51): if a+b+2*c+2*d ==100: count += 1 print count >>>45526 This is a slow method, finding a closed formula is better. I had to wait about 33 seconds for the answer, and if you increase 100 to 10000 or higher, you will wait longer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I found a better approach using this argument by mathmath333 in her previous post http://openstudy.com/study#/updates/55d8d229e4b05a670c275069 In addition we will use the theorem found here https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics). Consider two cases, when 'a' ,'b' are both even, and when 'a' ,'b' are both odd. You need not consider 'a' even and 'b' odd or vice a versa, since odd + even +even +even= odd != 100 case 1: a is even, b is even There exist nonnegative integers a', b' such that 2a' = a, 2b' = b, where a' can range from 0 to 50 , and b' can range from 0 to 50 2a' + 2b' + 2c + 2d = 100 a' + b' + c + d = 50 There are ( 50 + 4 1) choose (41) nonnegative solutions 53 choose 3 = 23426 case 2: a is odd, b is odd We want to end up with nonnegative integers a' , b ' starting with a' =0, b' = 0 so a = 2a'+ 1 , b = 2b' + 1 , is a good choice 2(a'+1) + 2(b'+1) + 2c + 2d = 100 2a' + 2b' + 2 + 2c + 2d = 100 2( a' + b' + c + d ) = 98 a ' + b ' + c + d = 49 There are ( 49 + 4 1) choose (41) nonnegative solutions 52 choose 3 = 22100 the total is 23426 + 22100 = 45526 this agrees with the python result

dan815
 one year ago
Best ResponseYou've already chosen the best response.0nice i really like that method

dan815
 one year ago
Best ResponseYou've already chosen the best response.0see if u can think about generalizations for this method like with k1*a+K2*b+k3*c+d...+z=k

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ill see what i can come up with, we might be able to write a completely dependant equation on the chosen constants k1,k2,k3...
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