We arrange 12-letter words having at our disposal five letters a, four letters b and three letters c

ziphumulegn 2022-07-01 Answered
We arrange 12-letter words having at our disposal five letters a, four letters b and three letters c. How many words are there without any block 5 × a, 4 × b and 3 × c. I need to use inclusion - exclusion principle. I counted all possible 12 letter words - 12 ! 5 ! 4 ! 3 ! . Then words with block of letters:
only a blocks - 8 7 ! 4 ! 3 !
only b blocks - 9 8 ! 5 ! 3 !
only c blocks - 10 9 ! 5 ! 4 ! .
Now I have to count all words, where block: a, b or a, c, or b, c appears together, but I don't now how.
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Answers (1)

Perman7z
Answered 2022-07-02 Author has 13 answers
Step 1
If we let N denote the total number of distinguishable arrangements of aaaaabbbbccc, A denote the set of permutations with five consecutive as, B denote the set of permutations with four consecutive bs, and C denote the set of permutations with three consecutive cs, then, by the Inclusion-Exclusion Principle, the number of arrangements with no block of five consecutive as or four consecutive bs or three consecutive cs is
| A B C | = N | A B C | = N ( | A | + | B | + | C | | A B | | A C | | B C | + | A B C | ) = N | A | | B | | C | + | A B | + | A C | + | B C | | A B C |
Step 2
You have correctly calculated N, |A|, |B|, and |C|.
| A B | We have five objects to arrange: aaaaa, bbbb, c, c, c. There are ( 5 3 ) ways to select three of the five positions for the three cs and 2! ways to arrange the remaining two distinct objects in the remaining two positions. Hence, there are
( 5 3 ) 2 ! = 5 ! 3 ! such arrangements.
Alternatively, arrange the three cs in a line. This creates four spaces in which we can place the block aaaaa, two between successive cs and two at the ends of the row.
c c c
Once we have placed the block aaaa, we have four objects in a row, which creates five spaces in which to place the block bbb, three between successive objects and two at the ends of the row. Hence,
| A B | = 4 5
Can you finish the argument by counting | A C | , | B C | , and | A B C | ?

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