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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-${}^{\text{n}}{{\text{C}}_{\text{r}}}$=$\dfrac{{n!}}{{r!n - r!}}$
So to select $3$ spaces of out $6$ spaces =${}^6{{\text{C}}_3}$
And the three vowels can be arranged in these three spaces in $3!$ ways.
$ \Rightarrow $ The total number of words formed=${}^6{{\text{C}}_3} \times 3! \times 5!$
$ \Rightarrow $ The total number of words
formed=$\dfrac{{6!}}{{3!6 - 3!}} \times 5! \times 3!$
$ \Rightarrow $ The total number of words formed=$\dfrac{{6!}}{{3!}} \times 5!$
On simplifying we get-
$ \Rightarrow $ The total number of words formed=$\dfrac{{6 \times 5 \times 4 \times 3!}}{{3!}} \times 5!$
$ \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 {}^n{P_r} = \dfrac{{n!}}{{n - r!}}$ Where n=total number of things and r=no. of things to be selected.
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In how many ways can the letters of the word 'APPLE' be arranged ?
|
| A.
| 720
| |
| B.
| 120
| |
| C.
| 60
| |
| D.
| 180
|
Answer & Explanation Answer: Option C Explanation: The word 'APPLE' contains 5 letters, 1A, 2P, 1L.and 1E. $$\therefore$$ Required number of ways = $$\frac{5 !}{(1 !) (2 !) (1 !) (1 !)}$$ = 60. How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed ?
|
| A.
| 40
| |
| B.
| 400
| |
| C.
| 5040
| |
| D.
| 2502
|
Answer & Explanation Answer: Option C Explanation: 'LOGARITHM' contains 10 different letters. Required number of words = Number of arrangements of 10 letters, taking 4 at a time = 10P4 = (10 * 9 * 8 * 7) = 5040. The value of 75P2 is :
|
| A.
| 2775
| |
| B.
| 150
| |
| C.
| 5550
| |
| D.
| None of these
|
Answer & Explanation Answer: Option C Explanation: 75P2 = $$\frac{75 !}{(75 - 2)!}$$ = $$\frac{75 !}{73 !}$$ = $$\frac{75 * 74 * (73 !)}{73 !}$$ = (75 * 74) = 5550. In how many ways can the letters of the word 'LEADER' be arranged ?
|
| A.
| 72
| |
| B.
| 144
| |
| C.
| 360
| |
| D.
| 720
|
Answer & Explanation Answer: Option C Explanation: The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R. $$\therefore$$ Required number of ways = $$\frac{6 !}{(1 !)(2 !)(1 !)(1 !)(2!)}$$ = 360. How many words with or without meaning, can be formed by using all the letters of the word, 'DELHI', using each letter exactly once ?
Answer & Explanation Answer: Option D Explanation: The word 'DELHI' contains 5 different letters. Required number of words = Number of arrangements of 5 letters, taken all at a time = 5P5 = 5 ! = (5 *4 *3 *2 *1) = 120. In how many different ways can the letters of the word 'RUMOUR' be arranged ?
|
| A.
| 180
| |
| B.
| 90
| |
| C.
| 30
| |
| D.
| 720
|
Answer & Explanation Answer: Option A Explanation: The word 'RUMOUR' contains 6 letters, namely 2R, 2U, 1M and 1U. $$\therefore$$ Required number of ways = $$\frac{6 !}{(2 !) (2 !) (1 !) (1!)}$$ = 180. How many arrangements can be made out of the letters of the word 'ENGINEERING' ?
|
| A.
| 277200
| |
| B.
| 92400
| |
| C.
| 69300
| |
| D.
| 23100
|
Answer & Explanation Answer: Option A Explanation: The word 'ENGINEERING' contains 11 letters, namely 3E, 3N, 2G, 2I and 1R. $$\therefore$$ Required number of arrangements = $$\frac{11 !}{(3 !) (3 !) (2 !)(2 !)(1 !)}$$ = 277200. How many words can be formed from the letters of the word 'SIGNATURE' so that the vowels always come together ?
|
| A.
| 720
| |
| B.
| 1440
| |
| C.
| 2880
| |
| D.
| 17280
|
Answer & Explanation Answer: Option D Explanation: The word 'SIGNATURE' contains 9 different letters. When the vowels IAUE are taken together, they can be supposed to form an entity, treated as one letter. Then, the letters to be arranged are SGNTR (IAUE). These 6 letters can be arranged in 6P6 = 6 ! = 720 ways. The vowels in the group (IAUE) can be arranged amongst themselves in
4P4 = 4 ! = 24 ways. $$\therefore$$ Required number of words = (720 * 24) = 17280. In how many different ways can the letters of the word 'SOFTWARE' be arranged in such a way that the vowels always come together ?
|
| A.
| 120
| |
| B.
| 360
| |
| C.
| 1440
| |
| D.
| 720
|
Answer & Explanation Answer: Option D Explanation: The word 'SOFTWARE' contains 8 different letters. When the vowels OAE are always together, they can be supposed to form one letter. Thus, we hdve to arrange the letters SFTWR (OAE). Now, 5 letters can be arranged in 6 ! = 720 ways. The vowels (OAE) can be arranged among themselves in 3 ! = 6 ways. $$\therefore$$ Required
number of ways = (720 * 6) = 4320. In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels ajways come together ?
|
| A.
| 120
| |
| B.
| 720
| |
| C.
| 4320
| |
| D.
| 2160
|
Answer & Explanation Answer: Option C Explanation: The word 'OPTICAL' contains 7 different letters. When the vowels OIA are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTCL (OIA). Now, 5 letters can be arranged in 5 ! = 120 ways. The vowels (OIA) can be arranged among themselves in 3 ! = 6 ways. $$\therefore$$ Required
number of ways = (120 * 6) = 720.
1 Answer. (e) The word SIGNATURE consists of nine letters comprising four vowels (A, E, I and U) and five consonants (G, N, R, T and S). When the four vowels are considered as one letter, we have six letters which can be arranged in 6P6 ways ie 6! ways. The vowels in the group (IAUE) can be arranged amongst themselves in 4P4 = 4 ! = 24 ways. Required number of words = (720 * 24) = 17280. If repetition is not allowed, we have 4 choices for the first letter, 3 choices for the second letter, and 2 choices for the third letter. Therefore, we can form 4*3*2 = 24 such “words". The word SIGNATURE has 9 different letters. The number of 3-letter words that can be
formed = 3! =720. Hence, the no. of 3 letter words formed from the word LOGARITHMS without repetition is 720. (ii) When repetition of letters is allowed, each place can be filled by any of the 5 letters in 5 ways. ∴ the required number of ways =(5×5×5)=125. ∴ the required number of ways = ( 5 × 5 × 5 ) = 125 . So, the
number of 3-letter words with or without meaning that can be formed using the letters of the word 'FLASH' is 60. Therefore, the number of words that can be formed using all the letters of the
word MONDAY, using each letter exactly once is 6×5×4×3×2×1=6! =720. So, the required number of 3-letter words =(5×4×3)=60. Solution : The number of different words that can be formed by using the four letters a,b,c,d, while the letter can be repeated is `4^4=256. ` Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. How many 3 letter words can be made if letters can be repeated?
ways to form a word with a repeated letter. Consequently, there are 24+18=42 distinguishable three letter words that can be formed with the letters of the word SERIES. Show activity on this post. If letters can be repeated as many times as you want, there are 6
options (A, B, C, D, E, or F) for the first letter, second letter, and third letter. Then 63=216 are the number of options for all three-letter-words.
The number of words formed from 'DAUGHTER' such that all vowels are together is 4320.
Is there a 5 letter word that uses all the vowels?
Unfortunately, there are no words in English that are made up entirely of vowels, so we will have to settle for the next best thing: a five-letter word containing four of them.
Number of ways when vowels are never together =120−48=72.
What 5 letter word uses the most vowels?
Top-Words with 5 Letters with mostly vowels. MIAOU. ... . ADIEU. ... . AUDIO. ... . AULOI. ... . LOUIE. ... . AUREI. ... . OURIE. ... . URAEI.. |
