My Math Forum How many combinations of 4 weights on a line?

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 April 21st, 2017, 04:03 AM #1 Newbie   Joined: May 2015 From: Berlin Posts: 9 Thanks: 1 How many combinations of 4 weights on a line? I'm not a mathematician - but here is a question that I would appreciate some help with: I have 4 types of weights that need to be hung on an unlimited amount of lines. The weights are:8 kg 7 kg 4 kg 3 kg There can be either 0, 1, 2, 3, 4, 5, or 6 weights hung on each line. The same weight can be on the line 0, 1, 2, 3, 4, 5, or 6 times (i.e. The same weight can be repeated). The weights on each the line are not allowed to exceed 20kg. How many combinations of weights are there? - and what are the combinations? I would be incredibly happy if anyone could either give me some answers. Thanks in advance! Last edited by Williams; April 21st, 2017 at 04:14 AM.
April 21st, 2017, 05:55 AM   #2
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Posts: 13,112
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Quote:
 Originally Posted by Williams The weights are:8 kg 7 kg 4 kg 3 kg There can be either 0, 1, 2, 3, 4, 5, or 6 weights hung on each line. The same weight can be on the line 0, 1, 2, 3, 4, 5, or 6 times (i.e. The same weight can be repeated). The weights on each the line are not allowed to exceed 20kg.
Your problem statement is a bit unclear...

If same weight is on line 6 times and maximum weight
is 20kg, then only 3kg can be used: is that correct?

What's a "line"?

May be easier if your problem was simplified a bit like:
using digits 0 to 6 up to a maximum of 6 times each,
how many arrangements are possible where the
sum =< 20 ?

Also: why are you complicating it by including weight 0?

April 22nd, 2017, 08:16 AM   #3
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Joined: May 2015
From: Berlin

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Thanks for your response Denis. You are right on all counts – I definitely could have written the question better. I’ll try the question again – phrased perhaps in a more applied mathematics way:
There are an unlimited number of washing lines which only can support a maximum of 20kg each.
There are also an unlimited amount of weights which need to be hung on the washing lines. The weights come in 4 ‘sizes’.
• 8kg
• 7kg
• 4kg
• 3kg
How many combinations of weights can I hang on the washing lines? And what are they?
---

If you were only allowed to have one type of weight per line the answer would be simple - 15:

1. No weights
2. 8 kg
3. 7 kg
4. 4 kg
5. 3 kg
6. 8 kg, 7 kg
7. 8 kg, 4 kg
8. 8 kg, 3 kg
9. 7 kg, 4 kg
10. 7 kg, 3 kg
11. 4 kg, 3 kg
12. 8 kg, 7 kg, 4 kg
13. 8 kg, 4 kg, 3 kg
14. 8 kg, 7 kg, 3 kg
15. 7 kg, 4 kg, 3 kg

The next stage: 8 kg, 7kg, 4 kg, 3kg wouldn’t work because it will be over 20kg.

But if the weights can be repeated – I’m lost and don’t know what to do next to work out the problem.

---

Quote:
 If same weight is on line 6 times and maximum weight is 20kg, then only 3kg can be used: is that correct?
Yes, that is right. The maximum amount of weights on a line would be 6.

Quote:
 Also: why are you complicating it by including weight 0?
I had 0 weights in the problem because having no weights on one of the washing lines is an option – but if this complicates matters then I’m happy to remove it and add it in at the end.

 April 22nd, 2017, 09:05 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,112 Thanks: 911 Looks like the easiest way to handle this is to use the 4 weights as the digits of a number. 333333 : 18 333334 : 19 333344 : 20 33333 : 15 33344 : 16 33337 : 19 33338 : 20 33344 : 17 33347 : 20 33444 : 18 34444 : 19 44444 : 20 Similarly for 4 weights, then 3, then 2 then 1. Notice that the numbers go from lowest to highest, and the digits are in style a=
 April 22nd, 2017, 10:49 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,091 Thanks: 1087 I get the following 57 combos that are 20 kg or less. $\begin{array}{c} \{\} \\ \{3\} \\ \{4\} \\ \{7\} \\ \{8\} \\ \{3,3\} \\ \{3,4\} \\ \{3,7\} \\ \{3,8\} \\ \{4,4\} \\ \{4,7\} \\ \{4,8\} \\ \{7,7\} \\ \{7,8\} \\ \{8,8\} \\ \{3,3,3\} \\ \{3,3,4\} \\ \{3,3,7\} \\ \{3,3,8\} \\ \{3,4,4\} \\ \{3,4,7\} \\ \{3,4,8\} \\ \{3,7,7\} \\ \{3,7,8\} \\ \{3,8,8\} \\ \{4,4,4\} \\ \{4,4,7\} \\ \{4,4,8\} \\ \{4,7,7\} \\ \{4,7,8\} \\ \{4,8,8\} \\ \{3,3,3,3\} \\ \{3,3,3,4\} \\ \{3,3,3,7\} \\ \{3,3,3,8\} \\ \{3,3,4,4\} \\ \{3,3,4,7\} \\ \{3,3,4,8\} \\ \{3,3,7,7\} \\ \{3,4,4,4\} \\ \{3,4,4,7\} \\ \{3,4,4,8\} \\ \{4,4,4,4\} \\ \{4,4,4,7\} \\ \{4,4,4,8\} \\ \{3,3,3,3,3\} \\ \{3,3,3,3,4\} \\ \{3,3,3,3,7\} \\ \{3,3,3,3,8\} \\ \{3,3,3,4,4\} \\ \{3,3,3,4,7\} \\ \{3,3,4,4,4\} \\ \{3,4,4,4,4\} \\ \{4,4,4,4,4\} \\ \{3,3,3,3,3,3\} \\ \{3,3,3,3,3,4\} \\ \{3,3,3,3,4,4\} \\ \end{array}$ Thanks from Williams
 April 23rd, 2017, 03:19 AM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,112 Thanks: 911 Agree. But can't come up with a formula. Wrote looper program. Thanks from Williams
 April 23rd, 2017, 05:21 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,112 Thanks: 911 Williams, if you're interested in writing a computer program for this silliness(!), then here's a sample: (I'll show the portion applicable to 5 weights) Open array size 4: a(4) a(1)=3 : a(2) = 4 : a(3) = 7 : a(4) = 8 loop a from 1 to 4 loop b from a to 4 loop c from b to 4 loop d from c to 4 loop e from d to 4 dowhile s < 21 s = a(a) + a(b) + a(c) + a(d) + a(e) print a(a), a(b), a(c), a(d), a(e), s Output: 3,3,3,3,3,15 3,3,3,4,4,16 3,3,3,3,7,19 3,3,3,3,8,20 3,3,3,4,4,17 3,3,3,4,7,20 3,3,4,4,4,18 3,4,4,4,4,19 4,4,4,4,4,20 Thanks from Williams Last edited by Denis; April 23rd, 2017 at 05:27 AM.
 April 23rd, 2017, 10:16 AM #8 Senior Member     Joined: Sep 2015 From: USA Posts: 2,091 Thanks: 1087 what I did was take the alphabet (0,3,4,7,8 ) and create all 6 tuples of it then select those tuples for which the sum of the elements is 20 or less. Then remove the 0's Mathematica is a real joy to work with once you get it, which took me years.... Thanks from Williams Last edited by romsek; April 23rd, 2017 at 10:19 AM.
April 23rd, 2017, 10:34 AM   #9
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Posts: 13,112
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Quote:
 Originally Posted by romsek Mathematica is a real joy to work with once you get it, which took me years....
Nothing will replace my love Miss UBasic...

 April 23rd, 2017, 11:07 PM #10 Newbie   Joined: May 2015 From: Berlin Posts: 9 Thanks: 1 This is all absolutely brilliant! Thank you for your work and help!!!

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