Combinatorics Question uOttawa (University of Ottawa)

Question

Consider the fourteen letters: $\text{A A A B B C C C C C D E E E}$ .

An ARRANGEMENT is a sequence using $\text{all}$ of these letters.

For the purposes of this question, a WORD is a sequence using $\text{some}$ of these letters.

a) How many arrangements can be made from these symbols?
b) How many arrangements contain ABBA as a subword (meaning those four letters in that consecutive order, somewhere in the arrangement).
c) How many words have all letters distinct?
d) How many arrangements have no two vowels consecutive?

Answer 1

(c) 

Alternative Explanation:

We need to create words using $A,B,C,D,E.$ So, choose $r$ elements from them, then we can permute them in $r!$ ways. This $r$ can range from $0$ to $5.$ So, answer is $\Sigma (5Cr)r!.$

(d) There are eight consonants B B C C C C C D and six vowels A A A E E E.

First arrange the consonants. This can be done in $8!/(5! \times 2!)$ ways. There are nine places before, between or after the consonants, where we may place at most one vowel. Choose six of these, and insert the vowels there. 

So, final answer: $[8!/(5! \times 2!) ] \times [ 9C6 ] \times [6!/(3! \times 3!)] $

Question Source: http://web5.uottawa.ca/mnewman/courses/2022-09.mat2348/test.2022-s.pdf 

Comments

@Deepak Poonia sir, in solution for [c], we are adding empty string in answer (when r=0 i.e., word of length 0). Is it correct?

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@Deepak Poonia Sir I didn’t get option c clearly

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@KG, Yes. 

BUT honestly speaking, it must be hinted in the question whether to include empty word or not.

For example, see the last line in the explanation of option C HERE. 

Anyway, Don’t Worry About It. 

@chokostar , added more explanation for (c).

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Thankyou sir, now that seems extremely simple!

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@Deepak Poonia sir, in option D is complement rule works 
like total-arrangements that have consecutive two vowels?

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@anirudhkumar18 , Indeed we can apply the complement rule. BUT that will be extremely tedious.

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Sir, in question (B) can't we arrange ABBA i.e. 4!/(2!)*(2!)?

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@Ashutosh Tripathy We want $ABBA$ substring, So, we can’t rearrange its letters.

For example, if we want $BDA$ substring in the permutation of $ABCDE$ then the valid permutations are $BDAEC, CBDAE$ etc But permutations like $CBADE, ABDCE$ etc are Invalid.  

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@Deepak PooniaGot it sir. Thank you..