2
$\begingroup$

How many ways can 2 a's, 2 b's and 8 c's be arranged so that there is a c on both sides of each a and b?

I'm really unsure of how to even begin tackling this. I would say treat each 'cac' and 'cbc' as their own object and permute, but that's not taking into account that the "before" c of a 'cac' combo could be the "after" c of a 'cbc' combo, aka "cbcac". How would I procede?

$\endgroup$
2
$\begingroup$

To ensure that there is a $c$ on both sides of each $a$ and each $b$, first place the eight $c$'s in a row. This leaves seven gaps between successive $c$'s.
$$c \square c \square c \square c \square c \square c \square c \square c$$ Select two of these seven gaps in which to place the $a$'s, then choose two of the five remaining gaps in which to place the $b$'s. This can be done in $$\binom{7}{2}\binom{5}{2}$$ ways.

$\endgroup$
  • $\begingroup$ what about ccaccacbcbcc $\endgroup$ – Uri Goren Mar 12 '16 at 22:22
  • 1
    $\begingroup$ @UriGoren: That is included in the count (see my answer, which has the same result). The arrangement you give had the choices 2nd, 4th for the a's, and 5th and 6th for the b's. $\endgroup$ – Rory Daulton Mar 12 '16 at 22:24
2
$\begingroup$

Use the stars and bars method here.

If you remove the a's and b's from the arrangement you just have 8 consecutive c's. By your restrictions, there cannot be an a or b to the left or to the right of all the c's, and the a's and b's cannot be next to each other.

Therefore we consider the 7 places between consecutive c's. We must place the 2 a's and the 2 b's in these 7 places, with each place taking at most one a or b. So choose 2 places for the a's: there are ${7 \choose 2}$ possibilities. Then choose 2 places for the b's from the remaining places: there are ${5 \choose 2}$ possibilities. Therefore the total number of arrangements is

$${7 \choose 2}{5 \choose 2}=21\cdot 10=210$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.