Advanced Techniques - Do not understand

For the problem in the "Advanced Techniques -- Bijections (Q6)", I do not understand how the answer gets (11 6) as the binomial coefficient, if n=15 and k=5. There are 15 seats and 5 people, but they can't sit adjacent to each other. I've always had a difficult time understand how to determine n and k. Any help would be greatly appreciated.

#Combinatorics

Easy Math Editor

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Comments

I think you're referring to this problem.

As described in the previous slide:

Let's imagine placing the 5 people (sitting on a chair) next to each other first, so there's $5!$ ways of doing so. Now you want to put one (unoccupied) chair between each of these 5 people.

$P \, E \, P \, E \, P \, E \, P \, E \, P$

( $P$ denotes a someone occupying a chair, while $E$ denotes an empty chair)

So far, you have only used up $5+4=9$ chairs. What about the other $15-9= 6$ empty chairs?

You can place any number of chairs in the red lines below:

${\color{#D61F06}{\text{\_}}} \, P \, {\color{#D61F06}{\text{\_}}} \, E \, P \, {\color{#D61F06}{\text{\_}}} \, E \, P \, {\color{#D61F06}{\text{\_}}} \, E \, P \, {\color{#D61F06}{\text{\_}}} \, E \, P \, {\color{#D61F06}{\text{\_}}}$

but as long as the total number of chairs use here is 6.

That is, you want to find the total number of positive integer solutions $(a,b,c,d,e,f)$ such that their sum is 6.

Can you see now why this is equivalent to placing 6 balls into 6 urns?

P.S.: Stars and bars is also relevant here if you chose not to use the bijection route.

Pi Han Goh - 3 years, 4 months ago

reference ?

André Hucek - 3 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$