Counting problem

- mathmath333

Counting problem

- chestercat

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- 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

hmm okay

- dan815

b can be all even numbers

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## More answers

- dan815

do you know how to solve the simpler version of this problem
a+b+c=100

- mathmath333

do i need to change 2b to d

- dan815

sry i mean 2b is all even numbers

- mathmath333

99C2

- dan815

yeah thats right

- mathmath333

i meant to use substitution as 2b=d

- 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

99C3

- dan815

i think you would solve
a+b+c=100
and divide by 2

- dan815

as all the odd possibilities would not possible

- mathmath333

u mean 99C2/2

- 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

How about calculatinb a+b1+b2+d=100 and substracting something

- mathmath333

?

- mathmath333

what do u mean

- dan815

ok back

- 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

no wait there are 101 WAYs actually to add upto 100
when u pick 2b=0

- dan815

|dw:1440274661135:dw|

- dan815

does 0 count as natural number though?

- mathmath333

No

- Zarkon

sometimes...some define it to include 0 and others don't

- anonymous

You could brute force it, write a python program to count it .

- dan815

okay then lets subtract that out

- mathmath333

0 is whole number

- mathmath333

No pen paper method i need

- Zarkon

http://mathworld.wolfram.com/NaturalNumber.html

- dan815

|dw:1440274943096:dw|

- dan815

2401

- dan815

do you want to check with brute force

- mathmath333

yea how do u brute force lol

- dan815

tbh its just writing what i did up there in a code lol

- dan815

i didnt really do it a very smart way summation is pretty bruteforcy

- dan815

the answer is definately close to (99 choose 2)/2

- dan815

because the odd + even combinations are not equal in this case there is going to be some fluctuation around that number

- mathmath333

http://www.wolframalpha.com/input/?i=solve+a%2B2b%2Bc%3D100%2Ca%3E0%2Cb%3E0%2Cc%3E0+over+integers

- 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

@kainui can you see a way to do this problem combinatorically

- 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

yea 2401 is correct

- 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

\(\large \color{black}{\begin{align}
& \normalsize \text{When a and c are even then}\hspace{.33em}\\~\\
& 2a'+2b+2c'=100\hspace{.33em}\\~\\
& \implies a'+b+c'=50\hspace{.33em}\\~\\
& \implies \dbinom{49}{2}\hspace{.33em}\\~\\
& \normalsize \text{When a and c are odd then}\hspace{.33em}\\~\\
& 2a'-1+2b+2c'-1=100\hspace{.33em}\\~\\
& \implies a'+b+c'=51\hspace{.33em}\\~\\
& \implies \dbinom{50}{2}\hspace{.33em}\\~\\
& \normalsize \text{Total}=\dbinom{49}{2}+\dbinom{50}{2}=2401\hspace{.33em}\\~\\
\end{align}}\)

- mathmath333

that pattern was incorrect

- 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

when n = 100,
1/2*100 -1 = 49
and 49^2 = 2401

- mathmath333

yea correct 49^2

- 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

u have to use stars and bars method
https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

- anonymous

interesting :)

- 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

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

yep

- mathmath333

the language given is wikipdia is technical , if u have any doubt u can go through stars and bars videos on youtubes

- anonymous

oh, do you have a link? or should i google

- 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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