Combinatorics and Sticks There are $n$ sticks lined up in a row and $k$ of them are to be chosen. 
a) How many choices are there?  


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*There are $n \choose k$ choices or $nCk$


b) How many choices are there if no two of the chosen sticks can be consecutively?


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*I think it's something like ${n-k+1} \choose k$   but I also think that whether there is an odd or even number of sticks matters... but I don't know how to show that.


c) How many choices are there if there must be at least $\ell$ sticks between each pair of chosen sticks? 
 A: Hint: Count the spaces between chosen sticks. Define $x_1$ to be the number of sticks before the first selected one, $x_2$ to be the number of sticks between the first and second selected ones, etc, and $x_{k+1}$ to be the number of sticks after the last selected stick.
Then $x_1 + x_2 + \dots + x_{k+1} = n - k$. Your constraints turn into constraints on the variables, i.e. for a) $x_i \geq 0$ for all i, for b) $x_1,x_{k+1} \geq 0$ and $x_2,\dots,x_k \geq 1$, etc.
Can you solve it now? (The $\binom{n+k-1}{k}$ should be useful...)
A: For b), you can first line up $n-k$ dots, representing the sticks NOT chosen.
This leaves $n-k+1$ gaps, counting the gaps outside the first and last dots, and there are $\binom{n-k+1}{k}$ ways to choose $k$ of these gaps in which to put the chosen sticks. $\;\;\;\;$ (OR use the method below, with $l=1$.)
For c), you can line up $k$ dots representing the sticks to be chosen, and then remove $(k-1)l$ blockers to be inserted at the end.  This leaves $n-k-(k-1)l$ sticks, so the number of ways to arrange the $k$ dots and $n-k-(k-1)l$ sticks in order is $\binom{n-(k-1)l}{k}$, the number of ways to choose the k places for the dots out of the total of $n-(k-1)l$ places.  (Now insert the blockers between consecutive pairs of dots.)
