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March 20th, 2019, 05:18 AM  #1 
Newbie Joined: Mar 2019 From: bangalore Posts: 7 Thanks: 0  Number of integer solutions..
How many integer solutions does this expression have: x1*x2*3*x4 = 770 So, 2*5*7*11 = 770 But the catch is that either one, two, or three of these variables can be 1 as well. So, 70*11*1*1 = 770 Similarly, 770*1*1*1 = 770. The way I proceeded with this question, was a follows: i) No "1": Number of cases = 4! = 24 ii) One "1": Number of cases = C(4,3)*4! = 96 iii) Two "1": Number of cases = C(4,2)*4!/2! =72 iv) Three "1": Number of cases = 4 So, total = 24 + 96 + 72 + 4 = 196 However, I believe (not 100% sure) that the correct answer is 256. Where am I going wrong Last edited by mohish; March 20th, 2019 at 05:37 AM. 
March 20th, 2019, 05:25 AM  #2 
Senior Member Joined: Feb 2010 Posts: 711 Thanks: 147  
March 20th, 2019, 05:36 AM  #3 
Newbie Joined: Mar 2019 From: bangalore Posts: 7 Thanks: 0 
Sorry...read it as 2⋅5⋅7⋅11

March 20th, 2019, 07:15 AM  #4  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Quote:
Your i) and iv) are correct Your ii) should be 144, your iii) should be 84 24 + 144 + 84 + 4 = 256 Ascending order 1: 1,1,1,770 2: 1,1,2,385 3: 1,1,5,154 4: 1,1,7,110 5: 1,1,10,77 ... 252: 154,5,1,1 253: 385,1,1,2 254: 385,1,2,1 255: 385,2,1,1 256: 770,1,1,1 Last edited by Denis; March 20th, 2019 at 07:33 AM.  
March 20th, 2019, 08:07 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
For one "1", for example, exactly two of the four primes must be replaced by their product, so the number of cases = C(4,2)*4! = 144.

March 20th, 2019, 08:24 AM  #6 
Newbie Joined: Mar 2019 From: bangalore Posts: 7 Thanks: 0  
March 21st, 2019, 06:51 AM  #7 
Newbie Joined: Mar 2019 From: bangalore Posts: 7 Thanks: 0  
July 6th, 2019, 04:07 AM  #8 
Newbie Joined: Jun 2016 From: Hong Kong Posts: 25 Thanks: 2 
That should be Stirling numbers of the second kind if you're distributing n elements into k parts $\frac{4!}{0!}\left\{ {4 \atop 4} \right\}=24$ $\frac{4!}{1!}\left\{ {4 \atop 3} \right\}=144$ $\frac{4!}{2!}\left\{ {4 \atop 2} \right\}=84$ $\frac{4!}{3!}\left\{ {4 \atop 1} \right\}=4$ 

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