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How many words can be formed from the letters of the word ‘DAUGHTER’ so that(i) The vowels always come together?(ii) The vowels never come together?
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    How many words can be formed from the letters of the word ‘DAUGHTER’ so that(i) The vowels always come together?(ii) The vowels never come together?
    How many different words can be formed of the letters of the word combine vowels always remain together?How many different words can be formed using the letters of the word combine?How many different words can be formed of the letters of the word combine so that vowels may occupy odd places?

Answer

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Hint: The word daughter has $8$ letters in which $3$ are vowels. For the vowels to always come together consider all the $3$ vowels to be one letter (suppose V) then total letters become $6$ which can be arranged in $6!$ ways and the vowels themselves in $3!$ ways.Complete step-by-step answer:
Given word ‘DAUGHTER’ has $8$ letters in which $3$ are vowels and 5 are consonants. A, U, E are vowels and D, G, H, T, R are consonants.
(i)We have to find the total number
of words formed when the vowels always come together.
Consider the three vowels A, U, E to be one letter V then total letters are D, G, H, T, R and V. So the number of letters becomes $6$
So we can arrange these $6$ letters in $6!$ ways. Since the letter V consists of three vowels, the vowels themselves can interchange with themselves. So the number of ways the $3$vowels can be arranged is $3!$
Then,
$ Rightarrow $ The total number of words formed will be=number of ways the $6$
letters can be arranged ×number of ways the $3$ vowels can be arranged
On putting the given values we get,
$ Rightarrow $ The total number of words formed=$6! times 3!$
We know $n! = n times left( n – 1 right)! times …3,2,1$
$ Rightarrow $ The total number of words formed=$6 times 4 times 5 times 3 times 2 times 1 times 3 times 2 times 1$
On multiplying all the numbers we get,
 $ Rightarrow $ The total number of words formed=$24 times 5 times 6
times 6$
$ Rightarrow $ The total number of words formed=$120 times 36$
$ Rightarrow $ The total number of words formed=$4320$
The number of words formed from ‘DAUGHTER’ such that all vowels are together is $4320$.

(ii)We have to find the number of words formed when no vowels are together.
Consider the following arrangement- _D_H_G_T_R
The spaces before the consonants are for the vowels so that no vowels come together. Since there are $5$ consonants so they can be
arranged in $5!$ ways.
There are $6$ spaces given for $3$ vowels. We know to select r things out of n things we write use the following formula-$^textntextC_textr$=$dfracn!r!n – r!$
So to select $3$ spaces of out $6$ spaces =$^6textC_3$
And the three vowels can be arranged in these three spaces in $3!$ ways.
$ Rightarrow $ The total number of words formed=$^6textC_3 times 3! times 5!$
$ Rightarrow $ The total number of words
formed=$dfrac6!3!6 – 3! times 5! times 3!$
$ Rightarrow $ The total number of words formed=$dfrac6!3! times 5!$
On simplifying we get-
$ Rightarrow $ The total number of words formed=$dfrac6 times 5 times 4 times 3!3! times 5!$
$ Rightarrow $ The total number of words formed=$120 times 5 times 4 times 3 times 2 times 1$
On multiplying we get,
$ Rightarrow $ The total number of words formed=$14400$
The total number of words
formed from ‘DAUGHTER’ such that no vowels are together is $14400$.

Note: Combination is used when things are to be arranged but not necessarily in order. Permutation is a little different. In permutation, order is important. Permutation is given by-
$ Rightarrow ^nP_r = dfracn!n – r!$ Where n=total number of things and r=no. of things to be selected.

How many different words can be formed out of the letters of the word ‘COMBINE’ so that:
(i) vowels always remain together ?

(ii) no two vowels are together ?

(iii) vowels may occupy odd places ?

Solution not provided.

Ans. (i) 720

       (ii) 1440

        (iii) 576

166 Views

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed.

Number of ways in which place (x) can be filled = 5

                                           m = 5

Number of ways in which place (y) can be filled = 4      (∵ Repetition
is not allowed)

                                           n = 4

Number of ways in which place (z) can be filled = 3      (∵ Repetition is not allowed)

                                           p.. = 3

∴ By fundamental principle of
counting, the total number of 3 digit numbers formed                                         = m x n x p.. = 5 x 4 x 3 = 60.

526 Views

A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

         
 

Event 1: A coin is tossed and the outcomes recorded.

                                Number of outcomes 

                                      m = 2

Event 2: The coin is tossed again and the outcomes recorded.

            Number of outcomes 

                                       n = 2

Event 3: The coin is tossed third time and the outcomes recorded.

           Number of
outcomes 

                                           p.. = 2

∴  By fundamental principle of counting, the total number of outcomes recorded     
                                               = 

548 Views

How many 3-digit odd numbers can be formed from the digits 1,2,3,4,5,6 if:
(a) the digits can be repeated (b) the digits cannot be repeated?

(a) Number of digits available = 6

Number of places [(x), (y) and (z)] for them = 3

Repetition is allowed and the 3-digit numbers formed are odd

Number of ways in which box (x) can be filled = 3 (by 1, 3 or 5 as the numbers formed are to be odd)

               m = 3
Number of ways of filling box (y) = 6                           (∴ Repetition is allowed)               n = 6

Number of ways of filling box (z) = 6                           (∵ Repetition is allowed)

              p.. = 6

∴  Total number of 3-digit odd numbers formed

                             = m x n x p.. = 3 x 6 x 6 = 108

(b) Number of ways of
filling box (x) = 3                     (only odd numbers are to be in this box )

                                   m = 3

Number
of ways of filling box (y) = 5                                (∵ Repetition is not allowed)

                           
  n = 5

Number of ways of filling box (z) = 4                                 (∵ Repetition is not allowed)

                     
       p.. = 4

∴     Total number of 3-digit odd numbers formed

                                  = m x n x p.. = 3 x 5 x 4 = 60.

231 Views

Given 5 flags of different colours, how many different signals can be generated if each signal requires use of 2 flags, one below the other?

Number of ways of finding a flag for place 1 = 5

                           m =
5

Number of remaining flags = 4

Number of ways of finding a flag for place 2 to complete the signal = 4

                     n = 4

∴ By fundamental principle of counting, the number of
signals generated                       = 

991 Views

How many 3-digit numbers can
be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is allowed.

Number of digits available = 5

Number of places for the digits = 3.

Number of ways in which place (x) can be filled = 5

                           m = 5

Number of ways in
which place (y) can be filled = 5    (∵  Repetition is allowed)

                            n = 5

Number of ways in which place (z) can be filled = 5    (∵ Repetition is allowed)

                             p.. = 5

∴ By fundamental principle of counting, the number of 3-digit numbers
formed.                           = m x n x p.. = 5 x 5 x 5 = 125

458 Views

How many different words can be formed of the letters of the word combine vowels always remain together?

This gives the number of possible arrangements as 3*2*1 = 6. In all there are 24*6 = 144 ways of arranging the letters. The required number of arrangements that satisfy the condition that vowels and consonants are not together is 144.

How many different words can be formed using the letters of the word combine?

64 words can be made from the letters in the word combine.

How many different words can be formed of the letters of the word combine so that vowels may occupy odd places?

In order that the vowels may occupy odd places, we first of all arrange any 3 consonants in even places in 4P3 ways and then the odd places can be filled by 3 vowels and the remaining 1 consonant in 4P4 ways. So, Required number of words = 4P3 × 4P4 = 24 × 24 = 576.
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