# Number of ways in which $3$ people can be selected from $n$ people sitting in a row and if no two of them are consecutive

Question:
If $P_n$ denotes the number of ways in which three people can be selected from $n$ people sitting in a row and if no two of them are consecutive and $P_{n+1}-P_n=15$, then find the value of $n$.

My attempt:
$P_n=\binom{\frac{n}{2}}{3}$ as three can be selected from $\frac{n}{2}$ people and other $\frac{n}{2}$ people cannot be selected.But I've reached a dead end while solving the problem. Please help me.
The answer is $8$.

• There is no question in this. Are you asking to find the value of $n$ such that $P_{n+1}-P_n = 15$? Or to prove that for all $n$, $P_{n+1}-P_n = 15$? (which is surely not true) – Mark Fischler Nov 25 '16 at 5:52
• Thanks I've edited it...it's to find the value of n – Sriram Cummaragunta Nov 25 '16 at 5:57
• Is the answer $n = 8$? – Shraddheya Shendre Nov 25 '16 at 6:02
• Your try does not work. It appears you think the three should all be in even or all be in odd positions, but for $n=6$ they could be $1,4,6$ – Ross Millikan Nov 25 '16 at 6:04
If you select any $3$ people out of the $n$ seating in a row, the people are divided into 4 regions (a rough sketch like the one below might help).
Let $x_1, x_2, x_3, x_4$ be the number of people in the $4$ regions. Then $$x_1 + x_2 + x_3 + x_4 = n-3$$ Now, here $x_2$ and $x_3$ can't be $0$, otherwise people selected will become consecutive. But $x_1$ and $x_4$ can be $0$. So, do the substitution $$x_2 = x_2^{'} + 1$$ $$x_3 = x_3^{'} + 1$$ Now as the least value of $x_2$ and $x_3$ is $1$, the least value of $x_2^{'}$ and $x_3^{'}$ is $0$. The new equation becomes $$x_1 + x_2^{'} + x_3^{'} + x_4 = n-5$$ Now the multinomial theorem is applicable as all the variables are non-negative. Hence, we get the value of $P_n$ as $\binom{n-2}{3}$. Solve the given equation now to get the value of $n$ as $8$.
First step is finding the closed form of $P_n$, which you've done wrong. Firstly you have to consider the number of all triples, which is $\binom{n}{3}$. Then subtract the 'bad' triples, which have at least two consecutive numbers, you can choose them in $(n-2)(n-1)$ ways, bacause for choosing the first of adjacent pair you can choose from the first $n-1$ numbers, then any of the remaining $n-2$ can be used for the third number. Now comes the tricky part. You have to subtract $n-2$ from the number of consecutive pairs, because if you have counted all triples from three adjacent numbers twice. So we finally we get $\binom{n}{3} - ((n-2)(n-1) - (n-2))$. As now you have closed form of $P_n$, substitute in the equation you have, simplify the expression, you'll get an equation for n, which you can easily solve.