# perms and combs qn (1 Viewer)

#### Masaken

##### Member

how would you do this with perms and combs? the worked solutions are using a probability tree and i think there's probably an easier way to go about this question, many thanks in advance

#### jimmysmith560

##### Le Phénix Trilingue
Moderator
The total number of ways can be given by $\bg_white \frac{6!}{2!\times 2!\times 2!}=90\:\text{ways.}$

Excluding arrangements where letters that are next to each other are identical from total arrangements results in the arrangements where no identical letters are next to one another. There are 3 situations where identical letters are next to one another:
• All 3 pairs together
• Only 2 pairs together
• Only 1 pair together
Having all three pairs next to one another can be expressed as A, A, B, B, C, C (just like in the question). In this case, the number of arrangements is $\bg_white 3!=6$

Having only 2 pairs next to one another can be expressed as C, A, A, B, B, C. The 2 pairs to be together can be selected in $\bg_white \binom{3}{2}=3\:\text{ways}$, with the third pair being the one where letters are not next to each other. The 2 chosen pairs can swap their places, causing 2 arrangements, each letter from the third pair can be placed in 3 positions, resulting in $\bg_white \binom{3}{2}\times 2\times \binom{3}{2}=18\:\text{ways}.$

Having only 1 pair together (i.e. the other 2 pairs of identical letters will not be next to one another) means that there are $\bg_white \binom{3}{1}=3\:\text{ways}$ to choose a pair of identical letters to be next to one another. On this basis, the number of placements for one pair of identical letters (for example, A) is $\bg_white \binom{5}{1}+\binom{5}{1}+1+1=12$. When considering all three pairs, this number becomes $\bg_white 3\times 12=36\:\text{ways}.$

The rest involves adding 6, 18 and 36 (resulting in 60) and subtracting this sum from 90, resulting in 30.

I hope this helps!