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 September 27th, 2018, 08:03 AM #1 Newbie   Joined: Sep 2018 From: India Posts: 2 Thanks: 0 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.
September 27th, 2018, 08:47 AM   #2
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 Originally Posted by mandeepkotti 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 ?

 September 27th, 2018, 09:37 AM #3 Newbie   Joined: Sep 2018 From: India Posts: 2 Thanks: 0 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.
 September 27th, 2018, 10:33 AM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 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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