Here's the question in my math homework this week.
Quote:
A shelf holds 12 books in a row. How many ways are there to choose 5 books so that no two adjacent books are chosen? [Hint: Represent the books that are chosen by bars and the books that are not chosen by stars. Count the number of sequences of five bars and seven stars so that no two bars are adjacent.]
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I could not figure out how to present this in one of the formulae we studied in this chapter. The hint was no help to me at all. I did a brute force method approach and counted the possible combinations.
Since the selections must be non-consecutive the longest we can wait to pick our first selection is book 4 and every other book after gives us:
4,6,8,10,12
Now working backwards on the shelf if our first book selected is 3 we get:
3,5,7,9,11
We can pick the same first four and choose 12 for our last books instead of 11 giving us:
3,5,7,9,12
With 3,5, and 7 picked we can also take 10 and 12 giving us:
3,5,7,10,12
With 3, and 5 we can take 8 as our third book giving us:
3,5,8,10,12
And our final pick with 3 as the first choice is:
3,6,8,10,12
Following an identical process and working through all the other options and tabulating the totals I came up with the following:
1 sequence with 4 as the first book selected
5 sequences with 3 as the first book selected
15 sequences with 2 as the first book selected
35 sequences with 1 as the first book selected
Total = 56 sequences
Is there a more elegant way to approach this problem? If we were talking about a shelf with 100 books and choosing 25 my solution would not be very practical.
A guy in my study group came up with C(8,3) which by formula is 8!/(3!5!) which does equal 56, so I suspect he's got the correct idea. I just can't see how he jumped to the formula from the given information since he didn't show any of his work. I've asked him to explain his process, but it's an online class and and I don't know how long it will take him to respond.
Anyone care to give me some help? Anyone, anyone? Buehler?