# Should we consider number of ways of choosing the first channel?

Poor Dolly's T.V. has only $$4$$ channels, all of them quite boring. Hence, it is not surprising that she desires to switch (change) channel after every one minute. Then find the number of ways in which she can change the channels so that she is back to her original channel for the first time after $$4$$ minutes.

In this question, I took the following approach: Let the channels be $$A, B,C,D$$ Without of loss of generality, let $$A$$ be the original channel(the one chosen first) Then total possible ways for $$1$$st switch,$$2$$nd switch, and $$3$$rd switch are $$3,2,2$$ respectively and after this, she switches to her original channel. So, No. of ways of switching the channels if A is chosen as the first channel=$$(3)(2)(2)=12$$ So,Total no. of ways$$=12+12+12+12=48$$ {Either of $$A/B/C/D$$ can be chosen as the first channel}

ANS:48

But the answer in my textbook is given to be $$12$$.Please point out the error in my solution.

Your error, is in caring about what the endpoints are. The start point, has already been chosen for Dolly when she turns the TV on. This also chooses the endpoint. She then has 3 other channels for the first choice, 2 she hasn't been to for the second choice, and 2 she can change to, that aren't her endpoint, for her last choice. This gives her 12 choices. The start and end point are fixed. I won't get into the book's possible errors, like double counting, because that wasn't asked.

• Can you please say whether my following conclusion is right/wrong? 1]Since they have mentioned about something called as original channel,we are taking that to be fixed one and hence don't consider the possibilities for the first channel. Mar 21 '19 at 16:07
• If you want to get the same answer as the book, it's a required assumption So yes.
– user645636
Mar 21 '19 at 16:11
• And can you suggest in which all cases should we consider the number of ways of choosing the starting point i.e,what modification in the question will give the answer as 48? Mar 21 '19 at 16:15
• When start and end channels aren't fixed.
– user645636
Mar 21 '19 at 16:19
• So,in these type of problems,if nothing is mentioned about starting point,then we have to assume that it is fixed? Mar 21 '19 at 16:23

The answer is given as 12 probably because the question asks for the number of ways to return to original channel for a particular starting channel and not for all possible original channels , therefore you need not multiply 12 by 4.
Now suppose we fix first channel to be A and we are left with B C D for first switch , one of them is selected which leaves us two switches for the second switch , as for the third switch we have the one left from the second switch and the one we chose for first switch , which resurfaces as a possible candidate .This gives us $$3×2×2=12$$ choices
However I would like to point out that this answer works only when you are not supposed to change back to original channel anytime between your 3 switches , we could have changed to A on second switch and then again to A at fourth switch , but somehow that's not acceptable from your answer '12' after all we are accepting other channels being repeated at alternative switches

Hint: What represents the factorial of a number?

You started to think correctly but you ended up counting the same event more than one time, in particular on the last switch.

Edit: I deleted a part of the answered based on a misinterpretation I did on the question.

• That would be watching A for 4 minutes which is forbidden by problem statement. This is a combinations with restricted repetition problem.
– user645636
Mar 20 '19 at 13:51
• @RoddyMacPhee: Thank you for your comment. I didn't pay enough attention on the question. I edited my answer to remove the wrong part of it. Mar 20 '19 at 14:42

The question asks to find the number of ways in which she can change the channels. The first channel is not her choice and so is the endpoint. therefore ans is $$12$$, not $$12\times4$$.