# Arranging letters with terms

This is a question I found in an old test while I was preparing for a test I have in a few days. The lecturer didn't really show us how to arrange with certain terms, so I need help with solving this.

Question 1 : How much arrangements are there for the word SOCIOLOGICAL when the letters AG have to be together.

Question 2 : Same question, only a different term : when all the letters A,O,I have to be together.

Thanks.

• For question 1, does it have to be that specific order, or is something like SOCIOLOICLGA acceptable? – Dennis Meng Sep 10 '13 at 15:57
• In Q1, difference of interpretation just means a factor of $2$. In Q2, meaning is quite unclear. Is it all the O's? Or do we want one of the permutations of AOI to appear at least once? – André Nicolas Sep 10 '13 at 16:03
• All the O's, and AOI isn't acceptable. – HaloKiller Sep 10 '13 at 16:15
• And yes, the order matters, so SOCIOLOICLGA is acceptable – HaloKiller Sep 10 '13 at 16:16

## 1 Answer

A common method for these sorts of problems to arrange the unrestricted characters, then figure out where you're allowed to put the rest. Think of the spaces between already-arranged letters as bins, in which to put the rest.

So for question 1: arrange the letters SOCIOLOICL in any way; there are $$\binom{10}{1,3,2,2,2}=\frac{10!}{1!\cdot 3!\cdot 2!\cdot 2!\cdot 2!}$$ ways to do this, since the multiplicities are 1(S), 3(O), 2(C), 2(I), and 2(L). Now, where can you put the A and the G? Well, of the 11 bins (9 between letters, 1 before the whole word, and 1 after the whole word), you must put them in the same one; and they can be arranged either AG or GA. So, there are $11\cdot2=22$ ways to position A and G, for a total of $$\binom{10}{1,3,2,2,2}\cdot22$$ ways.

For question 2: if both I's must be together and all three O's must be together, we can think of it this way: reduce our letters so that there is only one I and only one O. Arrange the resulting set of letters. Then expand the I to be II and the O to be OOO.

Reducing the letters yields SOCILGCAL; these can be arranged in $$\binom{9}{1,1,2,1,2,1,1}$$ ways. Then we expand the I and O, which can only be done in one way. So, this is the final answer here.

• All the A's,O's and I's are supposed to be in a sequence, for example: if we had 2 O's, 2 A's and 2 I's, AAOOII and IIAAOO are acceptable, whilst AOAIIO isn't. – HaloKiller Sep 10 '13 at 16:22
• But your string contains 1 A, 3 O, and 2 I. So, are you saying that we must have one of the strings AOOOII, AIIOOO, IIAOOO, IIOOOA, OOOAII, and OOOIIA occur someplace in your sequence, all together? – Nick Peterson Sep 10 '13 at 16:32
• He wanted to have A,G together, not apart. – NightRa Sep 10 '13 at 16:34
• @YonatanMarkman One moment. – Nick Peterson Sep 10 '13 at 16:41
• We can also see the pair or $A,G$ as an independant object, so we can arrage $11$ objects ($11!$), and then eliminate the duplicates (all that we devive), and then multiply by $2!$ to detonate the permutations inside $A,G$. – NightRa Sep 10 '13 at 16:41