Common approach for counting permutations is to choose the pattern, then choose the way to fill the pattern or vice versa.

For example:

How many $5$-letter words are there with exactly two vowels?

Start by choosing the places where the vowels will go. This is $2$ places out of $5$, so there are $10$ ways to do this. Then choose the vowels — $5^2$ — and the consonants — $21^3$. So the answer is $10 \cdot 5^2 \cdot 21^3$.

Often times, though, either the pattern or the elements are picked for us. For example,

Find the number of ways to rearrange three As and seven Bs if no two As can appear consecutively.

We can start with BBBBBBB and stick three A's in any of the eight spots between Bs. The answer's $\binom83$. In this problem we didn't have to choose the elements (A's). We were given them explicitly, so we only had to find spots for them.

Find the number of ways to select three distinct digits from the set $\{0, 1,2,\ldots,9\}$ if no two consecutive digits can be selected.

It's exactly the same problem as the one immediately precedes it. In this problem we must pick spots for triples where two are nonconsecutive integers, I think. The answer is $\binom83$.

My question is since we don't actually pick the non-consecutive integers, but only pick spots for them, does it mean we consider these triples given?


You can look at it this way:

There will be $7$ unchosen numbers (we don't know which), $\bullet\;\;\bullet\;\;\bullet\;\;\bullet\;\;\bullet\;\;\bullet\;\;\bullet$

The 3 chosen numbers must have been drawn from any 3 of the 8 gaps including the two ends:


  • $\begingroup$ Glad to see you back. $\endgroup$ – Shailesh May 11 '17 at 11:06
  • $\begingroup$ @Shailesh: Thanks, been a long time ! ): $\endgroup$ – true blue anil May 11 '17 at 17:04

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