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OpenStudy (mathmath333):
\(\large \color{black}{\begin{align}
& \normalsize \text{Find the number of natural number solutions of.}\hspace{.33em}\\~\\
& a+2b+c=100 \hspace{.33em}\\~\\
\end{align}}\)
OpenStudy (dan815):
hmm okay
OpenStudy (dan815):
b can be all even numbers
OpenStudy (dan815):
do you know how to solve the simpler version of this problem
a+b+c=100
OpenStudy (mathmath333):
do i need to change 2b to d
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OpenStudy (dan815):
sry i mean 2b is all even numbers
OpenStudy (mathmath333):
99C2
OpenStudy (dan815):
yeah thats right
OpenStudy (mathmath333):
i meant to use substitution as 2b=d
OpenStudy (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
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OpenStudy (mathmath333):
99C3
OpenStudy (dan815):
i think you would solve
a+b+c=100
and divide by 2
OpenStudy (dan815):
as all the odd possibilities would not possible
OpenStudy (mathmath333):
u mean 99C2/2
OpenStudy (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
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OpenStudy (mathmath333):
How about calculatinb a+b1+b2+d=100 and substracting something
OpenStudy (mathmath333):
?
OpenStudy (mathmath333):
what do u mean
OpenStudy (dan815):
ok back
OpenStudy (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
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OpenStudy (dan815):
no wait there are 101 WAYs actually to add upto 100
when u pick 2b=0