mathmath333
  • mathmath333
Counting problem
Mathematics
chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{Find the number of natural number solutions of.}\hspace{.33em}\\~\\ & a+2b+c=100 \hspace{.33em}\\~\\ \end{align}}\)
dan815
  • dan815
hmm okay
dan815
  • dan815
b can be all even numbers

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dan815
  • dan815
do you know how to solve the simpler version of this problem a+b+c=100
mathmath333
  • mathmath333
do i need to change 2b to d
dan815
  • dan815
sry i mean 2b is all even numbers
mathmath333
  • mathmath333
99C2
dan815
  • dan815
yeah thats right
mathmath333
  • mathmath333
i meant to use substitution as 2b=d
dan815
  • dan815
here is another way u can think about it a+b+b+c = 100 how would you solve this a+b+c+d=100
mathmath333
  • mathmath333
99C3
dan815
  • dan815
i think you would solve a+b+c=100 and divide by 2
dan815
  • dan815
as all the odd possibilities would not possible
mathmath333
  • mathmath333
u mean 99C2/2
dan815
  • dan815
yeah but let me think that might not technically be true we dont know if odd and even solutions that add to 100 is exactly equal or not
mathmath333
  • mathmath333
How about calculatinb a+b1+b2+d=100 and substracting something
mathmath333
  • mathmath333
?
mathmath333
  • mathmath333
what do u mean
dan815
  • dan815
ok back
dan815
  • dan815
okay well maybe we can go back to the stars and bars method after this but i think we can do it with just sums
dan815
  • dan815
no wait there are 101 WAYs actually to add upto 100 when u pick 2b=0
dan815
  • dan815
|dw:1440274661135:dw|
dan815
  • dan815
does 0 count as natural number though?
mathmath333
  • mathmath333
No
Zarkon
  • Zarkon
sometimes...some define it to include 0 and others don't
anonymous
  • anonymous
You could brute force it, write a python program to count it .
dan815
  • dan815
okay then lets subtract that out
mathmath333
  • mathmath333
0 is whole number
mathmath333
  • mathmath333
No pen paper method i need
Zarkon
  • Zarkon
http://mathworld.wolfram.com/NaturalNumber.html
dan815
  • dan815
|dw:1440274943096:dw|
dan815
  • dan815
2401
dan815
  • dan815
do you want to check with brute force
mathmath333
  • mathmath333
yea how do u brute force lol
dan815
  • dan815
tbh its just writing what i did up there in a code lol
dan815
  • dan815
i didnt really do it a very smart way summation is pretty bruteforcy
dan815
  • dan815
the answer is definately close to (99 choose 2)/2
dan815
  • dan815
because the odd + even combinations are not equal in this case there is going to be some fluctuation around that number
mathmath333
  • mathmath333
http://www.wolframalpha.com/input/?i=solve+a%2B2b%2Bc%3D100%2Ca%3E0%2Cb%3E0%2Cc%3E0+over+integers
mathmath333
  • mathmath333
a short research shows a+2b+c=10--->4! ways a+2b+c=12--->5! ways a+2b+c=14--->6! ways Can i somehow get a pattern
dan815
  • dan815
@kainui can you see a way to do this problem combinatorically
anonymous
  • anonymous
this would be the 'brute force' method Python code: count=0 for a in range(1,100): for b in range(1,100): for c in range(1,100): if a + 2*b+c ==100: count = count + 1 print count >>> 2401
mathmath333
  • mathmath333
yea 2401 is correct
anonymous
  • anonymous
a+2b+c=10--->4! ways a+2b+c=12--->5! ways a+2b+c=14--->6! ways you can derive a pattern from this
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{When a and c are even then}\hspace{.33em}\\~\\ & 2a'+2b+2c'=100\hspace{.33em}\\~\\ & \implies a'+b+c'=50\hspace{.33em}\\~\\ & \implies \dbinom{49}{2}\hspace{.33em}\\~\\ & \normalsize \text{When a and c are odd then}\hspace{.33em}\\~\\ & 2a'-1+2b+2c'-1=100\hspace{.33em}\\~\\ & \implies a'+b+c'=51\hspace{.33em}\\~\\ & \implies \dbinom{50}{2}\hspace{.33em}\\~\\ & \normalsize \text{Total}=\dbinom{49}{2}+\dbinom{50}{2}=2401\hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
that pattern was incorrect
anonymous
  • anonymous
edit* actually the pattern should be a+2b+c=10--->4^2 ways a+2b+c=12--->5^2 ways a+2b+c=14--->6^2 ways you can derive a pattern from this a+2b+c=n--->(1/2*n -1)^2 ways
anonymous
  • anonymous
when n = 100, 1/2*100 -1 = 49 and 49^2 = 2401
mathmath333
  • mathmath333
yea correct 49^2
anonymous
  • anonymous
thats a nice solution you have so how do you know that a' + b + c ' = 50 has (49 choose 2 ) ways to find positive solutions
mathmath333
  • mathmath333
u have to use stars and bars method https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
anonymous
  • anonymous
interesting :)
anonymous
  • anonymous
For any pair of positive integers n and k, the number of k-tuples of positive integers whose sum is n is equal to the number of (k − 1)-element subsets of a set with n − 1 elements. we had 50 is the sum , and 3 tuples so that means 49 choose 2
anonymous
  • anonymous
And then consider the case when a,b are both even or both odd. Note that a,c cannot be one even and one odd, because odd + even + even = odd
mathmath333
  • mathmath333
yep
mathmath333
  • mathmath333
the language given is wikipdia is technical , if u have any doubt u can go through stars and bars videos on youtubes
anonymous
  • anonymous
oh, do you have a link? or should i google
mathmath333
  • mathmath333
here is one of the links https://brilliant.org/discussions/thread/stars-and-bars/ or simply try google or youtube

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