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This is a question from permutation, word "ASSISA" can be arranged in how many ways such that all 'S' are togather?

karthik_personal
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    Ignore the fact that there are multiple $S$'s for now. Glue them all together. Now, you are just arranging the four letters $A\S I A$. – JMoravitz Apr 12 '21 at 15:28
  • Hi everyone, thanks for answers I have written the possibilities ; s' is block of s, AS'IA, S'AIA, IS'AA, AAIS', AAS'I, S'IAA, AIS'A, IAS'A. I have written 8 possibilities but answer is 12. Can anyone point missing possibilities? Also should two A's be considered different? – karthik_personal Apr 12 '21 at 16:15
  • "I have written the possibilities..." You should get out of the habit of doing this sooner rather than later. For problems as small as this, it is fine, but the whole point of a course in combinatorics is learning how to count without having to write them all out. – JMoravitz Apr 12 '21 at 16:17
  • As for writing these out... you should try to be more organized with how you do it. Try writing them out alphabetically. You are missing ASAI and IAAS from your list among others. By not being properly organized, it becomes very difficult to see what has or has not been skipped. – JMoravitz Apr 12 '21 at 16:18
  • As for "should two A's be considered different?" For this style problem, no. – JMoravitz Apr 12 '21 at 16:21
  • thanks @JMoravitz – karthik_personal Apr 12 '21 at 16:21

3 Answers3

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Imagine that the 3 "S"s form a block SSS. There's six letters in the word "ASSISA", let's denote them as: $$\text{_ _ _ _ _ _}$$

There are 4 ways to place the block SSS: $$\text{SSS _ _ _}$$ $$\text{_ SSS _ _}$$ $$\text{_ _ SSS _}$$ $$\text{_ _ _ SSS}$$

For the remaining 3 spaces, there's 3 ways to choose where "I" goes. Then, the two "A"s naturally go in the other two spaces. Therefore, there's $4*3 = \boxed{12}$ ways to arrange "ASSISA" with the "S"s all together.

Rebecca
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As JMoravitz started, you only need to place 4 letters A§IA. You can do that by choosing form 4 positions the letter I, then from 3 positions letter §. Then, two A´s are already in place. Therefore, 4*3=12

Albert Paradek
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How many ways can you arrange $n$ unique letters?

Now imagine labels $A_1, A_2, S_1, S_2, S_3, I$. That makes them all unique, so if you count the ways to arrange those, it is an overcount.

Now, how many ways are there to arrange $S_1. S_2, S_3$? $A_1, A_2$? Divide by those numbers to remove the redundancy.

RobertTheTutor
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