1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ For question 1, does it have to be that specific order, or is something like SOCIOLOICLGA acceptable? $\endgroup$ – Dennis Meng Sep 10 '13 at 15:57
  • $\begingroup$ 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? $\endgroup$ – André Nicolas Sep 10 '13 at 16:03
  • $\begingroup$ All the O's, and AOI isn't acceptable. $\endgroup$ – HaloKiller Sep 10 '13 at 16:15
  • $\begingroup$ And yes, the order matters, so SOCIOLOICLGA is acceptable $\endgroup$ – HaloKiller Sep 10 '13 at 16:16
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – HaloKiller Sep 10 '13 at 16:22
  • $\begingroup$ 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? $\endgroup$ – Nick Peterson Sep 10 '13 at 16:32
  • 1
    $\begingroup$ He wanted to have A,G together, not apart. $\endgroup$ – NightRa Sep 10 '13 at 16:34
  • 1
    $\begingroup$ @YonatanMarkman One moment. $\endgroup$ – Nick Peterson Sep 10 '13 at 16:41
  • 1
    $\begingroup$ 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 $A,G$. $\endgroup$ – NightRa Sep 10 '13 at 16:41

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.