Solutions to an X1+X2...+X5=21 EQN

Havie

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Hello again math world!
I am stumped on part d) of a question concerning a very particular set of rules.
heres the Q:
Q15.PNG


I'll also attach my attempt at an answer if anyone cares to see

I awkwardly arrive at the right answer of 106 on pg2but feel like i did so in the wrong way.


the idea is that were putting balls into distinct boxes kinda like pigeon hole principle.

since X3 has to have at least 15 and questionably X2 has to have at least 1
we either subtract 16 (or 15) from the total balls (21)

in my work I figured we'd have 5 total balls left then to distribute into these boxes.

first I tried separating into cases and subtracting the double counting of when x1=0 and x2=0 but this got weird and gave me an answer of 116.
Also in this step I realized there could never be 3 X1s and 3 X2s this way because thats 5-6 balls then for your choices.

then on PG2 I tried separating them out again into cases and not including the instances where x2 would be 1, simply because it SHOULD already be 1 because in the beginning i subtracted 15+1 from 21 to handle x2 being at least 1....
so I took those out and got 97..... realizing i was 9 short from the books answer... i looked and noticed if i re-added those choices where x2 would be 1 back in... i got 106.....
which just made absolutely no sense,
that got me thinking... there shouldnt even be an instances where x2=0.... so if i took those out,
i only get 45... which is way farther from the answer, and im now im totally stumped on how to handle this eqn.

Looking for an explanation of where i went wrong please and ty!
 
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nvm i got it!

gotit.jpg

its all about not subtracting that 1 X2 ball in the beginning, and handling it in the cases.
 
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Hello again math world!
I am stumped on part d) of a question concerning a very particular set of rules.
heres the Q:
View attachment 10499

I know that you have done this the long way & correctly.
I just wanted you to see THIS WEBPAGE

​There you will see the term \(\displaystyle 106x^{21}\) which answers part d).
 
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I know that you have done this the long way & correctly.
I just wanted you to see THIS WEBPAGE

​There you will see the term \(\displaystyle 106x^{21}\) which answers part d).

darn. that makes absolutely zero sense to me, i have no idea whats going on that webpage. >.<
 
darn. that makes absolutely zero sense to me, i have no idea whats going on that webpage. >.<
That is solutions by generating functions.
Suppose the question were: How many natural number solution are there for:
\(\displaystyle u+v+w+x=14\) subject to these conditions:
\(\displaystyle u\ge 5,~v\le 10\) and \(\displaystyle w\text{ is even while }x\text{ is odd}~?\)

To solve that is nightmare if you do cases. So lets use a generating function.
\(\displaystyle (\sum\limits_{k = 5}^{14} {{t^k}} )(\sum\limits_{k = 0}^{10} {{t^k}} )(\sum\limits_{k = 0}^7 {{t^{2k}}} )(\sum\limits_{k = 1}^7 {{t^{2k - 1}}} )\) Look at the expansion here.

The term \(\displaystyle 55t^{14}\) tell us that there are fifty-five solutions with those conditions.
 
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