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September 27th, 2018, 08:03 AM   #1
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Lightbulb Permutations and Combinations Problem

There are n number of chocolates. We can pick either 1 or 3 chocolates at once.
In how many ways can we pick all the chocolates?

The n variable can have any value as 30,100,50 etc.
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September 27th, 2018, 08:47 AM   #2
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Originally Posted by mandeepkotti View Post
There are n number of chocolates. We can pick either 1 or 3 chocolates at once.
In how many ways can we pick all the chocolates?

The n variable can have any value as 30,100,50 etc.
30,100,50 ??? Do you mean 30,100,500?

If n=1 or 2, then ?
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September 27th, 2018, 09:37 AM   #3
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Smile

If n=1: There can be one possible way.
If n=2: There are two possible ways. One on the first pick, and another one on second.
if n=3: There can be 4 combinations. Either all chocolates at once or one by one.

I hope this gives an idea about the problem.
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September 27th, 2018, 10:33 AM   #4
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The problem is given $n$ find $j, ~k \ni 3j + k = n$

fixing $j$ we see that $k = n - 3j$ i.e. it is fixed for a given $j$

We also note that $0 \leq j \leq \left \lfloor \dfrac{n}{3} \right \rfloor$

So a given $n$ has $ \left \lfloor \dfrac{n}{3} \right \rfloor+1$ different combinations of 1's and 3's that sum to it.
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