Kenneth Rosen Edition 6th Exercise 5.3 Question 31 (Page No. 362)

Question

The english alphabet contains 21 consonants and five vowels.How many strings of six lowercase letters of the English alphabet contain

(b)Exactly two vowels

(d)At least two vowels

For (b) part I solved it like choose 2 vowels from 5 in $\binom{5}{2}$ ways and remaining 4 consonants in $\binom{21}{4}$ ways and permute this string in 6! ways. So total -> $\binom{5}{2}$ * $\binom{21}{4}$ *6!

and for (d) part

I did it by complimentary counting and subtracted case of having atmost 1 vowel from all combinations of strings of 6 letters.

and this was

$26^6$ - $(21^{6} - (\binom{5}{1}*\binom{21}{5}*6!))$

and both of my answers don't match with the key in Rosen. Please let me know where I am wrong.

Comments

You made an assumption no latter is repeated.

For unlimited repeatation :

(1) 21⁵ *6*5 i.e. consonants and vowels can be reaped with restriction consonants five times and vowels two times.

(2) you need to do total - atleast one (otherwise you loose equal to two from atleast 2 )

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I also think no letters should be repeated here

because only 26 letters they assume

So, ans for

b)$6\times 5\times 21\times 20\times19 \times18 \times 6!$

isnot it?

Answer 1

It is considering repetitions to be allowed. In that case,

For b)

Two characters have 5 options(vowels) = 5²

4 characters have 21 options(consonants) = 21⁴

They can be arranged in 6!/(4!*2!)

Thus, the final answer = 5² * 21⁴ * 6!/(4!*2!)

For d)

Total number of arrangements - (number of arrangements with no vowel + number of arrangements with 1 vowel)

Total number of arrangements = 26⁶

Number of arrangements with no vowel = 21⁶ *(6!/6!)

Number of arrangements with 1 vowel = 5*21⁵ * (6!/5!)

Comments

is answer matching ?

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yes

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Yes for both parts, answers are correct.

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but

$\frac{6!}{4!2!}$ means we have to choose 1 vowel for all 2 positions and 1 consonant for all 4 positions and that should not be the case

right?

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no it means from 6 objects we consider all permutations in which 4 are of one kind and 2 are of another.

Similar to number of permutations of the word EVERGREEN

So here we have 4 E's 2 R's ,V,G,N = $\frac{9!}{4!2!}$

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See take a,e for vowel, i,j,k,l for consonant

now,$\binom{5}{1}^{2}\binom{21}{1}^{4}\times 6!$ will be the case

if every element is different , why do we divide?

Answer 2

Considering Vowels and Consonants can be repeated, then total 6 letter strings that can be formed :-

$$C(6,2) * 5 * 5 * C(4,4) * 21 * 21 * 21 * 21$$

Answer 3

There are 6 locations and out of these 6 locations two vowels can be chosen in 6C2 ways.

In the two locations of the vowels, vowels can be chosen in 5*5 ways.

now remaining 4 locations can be filled with 21*21*21*21 ways.

So, no of strings of six lowercase letters of the English alphabet that  contain exactly two vowels =6C2 * 5*5 * 21*21*21*21 =72930375

Answer 4

$b)$ Say, it can be a repeated characters string or non repeated character string or combination of repeated and non repeated character string

Firstly when always vowels are repeating

Case $1$: Say all vowel are repeating and all non vowel are repeating ${\color{Blue}{----}}{\color{Orange} {--}}$=$\binom{5}{1}\binom{21}{1}\frac{6!}{4!2!}=1575$

Case $2$: Say, all vowel are repeating and 1 consonant is repeating 3 times${\color{Blue}{---}}{\color{Cyan} -}{\color{Orange} {--}}$$\binom{5}{1}\binom{21}{1}\binom{20}{1}\frac{6!}{3!2!}$

                                                                                                                    $=5\times 21\times 20\times \frac{6\times 5\times 4}{2}=126000$

Case $3$: Say, 2 consonant are repeating 2 times=${\color{Blue} {--}}{\color{Teal} {--}}{\color{Orange} {--}}$$\binom{5}{1}\times \binom{21}{2}\frac{6!}{2!\times 2!\times 2!}=94500$

Case $4$: Say, no consonant are repeating$\binom{5}{1}\times \binom{21}{4}\frac{6!}{2!}=10,773,000$

Case $5$: Say, if there are three consonant , where $2$ repeating and $2$ non repeating ${\color{Blue} {--}}{\color{Magenta} {-}}{\color{DarkGreen} {-}}{\color{Orange} {--}}$

$\binom{5}{1}\times \binom{21}{1}\times \binom{20}{1}\times \binom{19}{1}\times \frac{6!}{2!\times 2!}=7,182,000$

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All these 5 cases when no vowels are repeating

Case $1$: Say all vowel are non repeating and all non vowel are repeating$\binom{5}{2}\binom{21}{1}\frac{6!}{4!}=6300$

Case $2$: Say, no vowel are repeating and 1 consonant is repeating 3 times=$\binom{5}{2}\binom{21}{1}\binom{20}{1}\frac{6!}{3!}=504000$

Case $3$: Say, $2$ consonant are repeating $2$ times=$\binom{5}{2}\binom{21}{2}\frac{6!}{2!2!}=378000$

Case $4$: Say, no consonant are repeating=$\binom{5}{2}\binom{21}{4}\times 6!=43,092,000$

Case $5$: Say, if there are three consonant , where $2$ repeating and $2$ non repeating=$\binom{5}{2}\times \binom{21}{1}\binom{20}{1}\binom{19}{1}\frac{6!}{2!}=28,728,000$

Now , summation of all these will be 90,883,800