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Can you help me with this problem of combinatorics?

There are $12$ lights in a row enumerated from $1$ to $12$. The lights $3$, $7$ and $11$ are switched on and the others are off. One of the lights can be switched on if at least one of its neighbours is switched on and the lights can't be switched off. In how many ways could you switch on all the lights?

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    If I understand correctly, step $0$ is the has the set of possibilities $L = \{2,4,6,8,10,12\}$, and so there are 6 ways of turning on $4$ lights? – 2017-01-15
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    @O.VonSeckendorff Yes, step $0$ has the set of possibilities $L=\{2,4,6,8,10,12\}$. – 2017-01-15

1 Answers 1

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METHOD 1

Lights numbered as follows (orange indicates already switched on): $$1, 2, \color{orange}3,4,5,6,\color{orange}7,8,9,10,\color{orange}{11},12$$

Note that possible subsequences (with given names), which must be sequential but not necessarily consecutive, are:

$Z=12$

$P,Q=2, 1$

$A,B,C= (4,5,6), (4,6,5),(6,5,4),(6,4,5)$

$D,E,F= (8,9,10),(8,10,9),(10,9,8),(10,8,9)$

There are 9 spaces to be filled (order of switching on). Think of this as arranging letters Z,P,Q,A,B,C,D,E,F where subsequences above must hold.

Start with $9$ spaces.

First place Z. There are $\color{blue}9$ ways.

This leaves $8$ spaces.

Next place $P,Q$ s.t. they are sequential but not necessarily consecutive.
Number of ways = $$\sum_{p=1}^7\sum_{q=p+1}^8 1=\sum_{q=2}^8\sum_{p=1}^{q-1}1 =\sum_{q=2}^8\binom {q-1}1=\binom 82=\color{blue}{28}$$

This leaves $6$ spaces.

Now place $A,B,C$ s.t. they are sequential but not necessarily consecutive.
Number of ways= $$\sum_{a=1}^4\sum_{b=a+1}^5\sum_{c=b+1}^6 1 =\sum_{c=3}^6\sum_{b=2}^{c-1}\sum_{a=1}^{b-1}1 =\sum_{c=3}^6\sum_{b=2}^{c-1}\binom{b-1}1 =\sum_{c=3}^6\binom{c-1}2 =\binom{6}3=20$$

Multiply this by $4$ (as there are $4$ possibilities for $A,B,C$) to give $\color{blue}{80}$.

This leaves $3$ spaces.

There is only $1$ way to places $D,E,F$ they are sequential (there are only $3$ spaces left for $3$ letters and they may or may not be consecutive). Multiply this by $4$ (as there are $4$ possibilities for $D,E,F$) to give $\color{blue}4$.

Total number of ways = $\color{blue}{9\cdot 28\cdot 80\cdot 4}=\color{red}{80640}$

METHOD 2

(This method can be deduced directly or from the resultant binomial coefficent of the analysis above. It is similar to the approach noted in @Tad's comment below.)

Another way would be to think of these as the number of ways to arrange letters $ZEEFFFGGG$, which is $$\binom 91\binom 82\binom 63\binom 33=\binom 9{1,2,3,3}$$ Now count the number of ways for:

  • $P,Q$ to be inserted sequentially into the two positions of $E,E$. (only $\color{purple}1$ way)

  • $A,B,C$ to be inserted sequentially into the three positions or $F,F,F$ ($\color{purple}4$ ways, as there are $4$ possible arrangments of $A,B,C$)

  • $D,E,F$ to be inserted sequentially into the three positions or $G,G,G$ ($\color{purple}4$ ways, as there are $4$ possible arrangments of $D,E,F$)

Hence the total number of ways of switching on all lights is given by $$\color{purple}1\cdot \color{purple}4\cdot \color{purple}4\cdot \binom 9{1,2,3,3}=\color{red}{80640}$$

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    In your last you want $D,E,F$ instead of $P,Q,R$, but a good answer. – 2017-01-15
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    @RossMillikan - Thanks (corrected), and thanks! :) – 2017-01-15
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    Nice solution - here's how I'd organize it. There are 4 intervals of lights, having lengths 1, 2, 3, 3. These can be shuffled together arbitrarily, so if we first pick which of the 9 slots are used for which interval, we get $\binom{9}{3,3,2,1}=5040$ choices. Now @hypergeometric points out there are 1, 1, 4, and 4 legal orders in which these intervals can be (independently) filled in, so the final count is $5040\cdot16=80640.$ – 2017-01-15
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    @Tad - Thanks! Nice multinomial approach. I have inserted an addendum to include another method based on the resultant binomial coefficients in the original anlaysis, which is similar to the approach you noted. – 2017-01-16