The given word, 'DAUGHTER' contains 3 vowels A, U, E and 5 consonants, D, G, H, T, R.Case (i) When all vowels occur together :Let us assume (AUE) as a single letter.Then, this letter (AUE) along with 5 other letters can be arranged in
6P6=(6!) ways =
(6×5×4×3×2×1) ways= 720 ways.These 3 vowels may be arranged among themselves in 3
!=6 ways.Hence, the required number of words with vowels together= (6
!)×(3
!)=(720×6)=4320.Case (ii) When all vowels do not occur together.Number of words formed by using all the 8 letters of the given word=
8P8=8
!=(8×7×6×5×4×3×2×1)=40320.Number of words in which all
vowels are never together = (total number of words) - (number of words with all vowels together)=(40320−4320)=36000.
Question 1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Solution:
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Question 1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Question 2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur
together?
Question 3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the
committee consists of: (i) exactly 3 girls ? (ii) atleast 3 girls ? (iii) atmost 3 girls ?
Question 4. If the different
permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?
Question 5. How
many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7, and 9 which are divisible by 10, and no digit is repeated?
Question 6. The English
alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Question 7. In an
examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
Question 8. Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Question 9. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Question 10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3
students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?
Question 11. In how many ways can the letters of the word
ASSASSINATION be arranged so that all the S’s are together?
How many different words can be formed from the letters of the word daughter?
How many different words can be formed from the word daughter ending and beginning are consonants?
How many different eight letter words can be formed out of the word daughter so that?
In the word “DAUGHTER”, there are 3 vowels, namely, A, U, E, and 5 consonants, namely, D, G, H, T, R.
The number of
ways of selecting 2 vowels out of 3 = 3C2 =
= 3.
The number of ways of selecting 3 consonants out of 5 = 5C3
=
=10.
Therefore, the number of combinations of 2 vowels and 3 consonants is 3 x 10 = 30.
Now, each of these 30 combinations has 5 letters which can be arranged among themselves in 5! ways.
Therefore, the required number of different words is 30 x 5! = 30 x 120 = 3600
Question 2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Solution:
There are 8 different letters in the word “EQUATION”, in which there are 5 vowels,
namely, A, E, I, O and U and 3 consonants, namely, Q, T and N. Since the vowels and consonants have to occur together, assume all vowels as one object (AEIOU) and all consonants as another (QTN). So, there are 2 objects now.
Then, permutations of these 2 objects taken all at a time = 2P2 = 2! =2.
Within vowel group, we have 5! Possible permutations and 3! permutations within the consonants group.
Hence, the required number of permutations = 2! x 5! x 3! = 1440.
Question 3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls ? (ii) atleast 3 girls ? (iii) atmost 3 girls ?
Solution:
(i) We have to select
7 members from 13 (9 boys + 4 girls).
If there are exactly 3 girls in each combination, then
The number of ways of selecting 3 girls out of 4 = 4C3 =
= 4
The number of ways of selecting 4 boys out of 9 = 9C4 =
= 126
Hence, total number of ways to form committee with exactly 3 girls = 126 x 4 = 504
(ii) Since, the team has to consist of at least 3 girls, the team can consist of
3 girls and 4 boys, or 4 girls and 3 boys.
Case 1: The team can consist of 3 girls and 4 boys:
So,
from(i), we have,
Total number of ways to select 3 girls and 4 boys = 126 x 4 = 504
Case 2: The team can consist of 4 girls and 3 boys:
Total number of ways to select 4 girls and 3 boys = 4C4 x 9C3 = 1 x 84 = 84
Therefore, total number of ways to form a committee with at least 3 girls = 504 + 84 = 588
(iii) Since, the team has to consist of at most 3 girls, the team can consist of
3 girls and 4 boys, or 2 girls and 5 boys, or 1 girl and
6 boys, or 7 boys
Case 1: The team can consist of 3 girls and 4 boys
Total number of ways to select 3 girls and 4 boys = 126 x 4 = 504
Case 2: The team can consist of 2 girls and 5 boys
Total number of ways to select 2 girls and 5 boys = 4C2 x 9C5 =
= 126 x 6 = 756
Case 3: The
team can consist of 1 girl and 6 boys
Total number of ways to select 1 girl and 6 boys = 4C1 x 9C6 =
= 84 x 4 = 336
Case 4: The team can consist of 7 boys
The number of ways of selecting 7 boys out of 9 = 9C7 =
= 36
Therefore, total number of ways to form a committee with at most 3 girls = 504 + 756 + 336 + 36 = 1632
Question 4. If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?
Solution:
In a dictionary, words are in ascending order. Hence,
for this scenario, all the words starting with A will come before first word starting with E.
To get the number of words starting with A, we fix the letter A at the extreme left position and then rearrange the remaining 10 letters taken all at a time. These 10 letters includes I and N 2 times.
Hence, the number of words starting with A =
= 907200.
Hence, 907200 words are there in the list before the first word starting with E.
Question 5. How many 6-digit numbers can be formed from the
digits 0, 1, 3, 5, 7, and 9 which are divisible by 10, and no digit is repeated?
Solution:
We know that, a number is divisible by 10 if it has 0 at its unit place. Hence, 0 will be fixed at last place in a number.
The remaining 5 places can be filled by any digit 1, 3, 5, 7 or 9. These 5 digits can be arranged at 5 places in 5P5 = 5! ways.
Hence, 6-digit numbers that can be formed which are divisible by 10 = 5! = 120.
Question 6. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Solution:
We have to select 2 vowels from 5 and 2 consonants from 21.
The number of ways to select 2 vowels from 5
= 5C2 =
= 10
The number of ways to select 2 consonants from 21 = 21C2 =
= 210
Permutations of these 4
alphabets taken all at a time = 4P4 = 4! = 24.
Hence, number of words can be formed with 2 different vowels and 2 different consonants = 10 x 210 x 24 = 50400
Question 7.
In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
Solution:
Since, it is required to attempt at least 3 questions from each part, questions selection can be
Case 1: (3, 5), or (b)(4, 4), or (c) (5,
3) ……(Part I questions, Part 2 questions)
Part I consists of 5 questions and Part II consists of 7 questions.
So, Number of ways to select 3 questions from Part I = 5C3 =
= 10
Number of ways to select 5 questions from Part II = 7C5 =
= 21
Hence, total number of ways for this type of question selection = 10 x 21 = 210
Case 2: Number of ways to select 4 questions from Part I = 5C4 =
= 5
Number of ways to select 4 questions from Part II = 7C4 =
= 35
Hence, total number of ways for this type of question
selection = 5 x 35 = 175
Case 3: Number of ways to select 5 questions from Part I = 5C5 = 1
Number of ways to select 3 questions from Part II = 7C3 =
= 35
Hence, total number of ways for this type of question selection = 1 x 35 = 35
Therefore, a student can select the questions in
210 + 175 + 35 = 420 ways.
Question 8. Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Solution:
We have to select 5 cards from 52 cards. If there is exactly one king in each combination, then
Case 1: We have to
select 1 king card from 4 king cards
Number of ways to select king card = 4C1 =
= 4
Case 2: We have to select 5 – 1 = 4 cards from remaining 52 – 4 = 48 cards
Number of ways to select remaining 4 cards = 48C4 =
= 194580
And, hence required total number of 5 card combinations = 4 x 194580 = 778320.
Question 9. It is required to seat 5 men and 4 women in a row so
that the women occupy the even places. How many such arrangements are possible?
Solution:
There are 5 men and 4 women. Women occupy the even places. Hence, the possible seating arrangement of men and women will be MWMWMWMWM where M denotes Man and W denotes Woman.
Now, 4 women can occupy 4 even places in 4P4 = 4! ways.
And, 5 men can occupy 5 odd places in 5P5 = 5! ways.
Hence, Total number of such possible arrangements:
5!
x 4! = 5 x 4 x 3 x 2 x 1 x 4 x 3 x 2 x 1 = 2880.
Question 10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can
the excursion party be chosen?
Solution:
According to those 3 students who decide to either join all of them or join none of them, there are 2 cases for excursion party member selection:
Case 1: Do not select those 3 students.
Hence, select 10 students from remaining 25 – 3 = 22 students.
The number of ways to do this = 22C10 = 646646
Case 2: Select those 3 students.
This means, we are short of 7 students and
these can be selected from remaining 25 – 3 = 22 students.
The number of ways to do this = 22C7 x 3C3 = 170544
Hence, required number of ways to choose excursion party:
646646 + 170544 = 817190
Question 11. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Solution:
Word “ASSASSINATION” has 4 S. We have to place all S’s together. Hence, we will assume group of 4 S as a single object. This single object together with 9 remaining letters (objects) will be counted as 10 objects to be arranged. These include 3 A’s, 2 I’s and N’s and 1 T and O.
So, required number of ways =
= 151200
How many different words can be formed from the letters of the word daughter?
The number of words formed from 'DAUGHTER' such that all vowels are together is 4320.
How many different words can be formed from the word daughter ending and beginning are consonants?
1 Answer. There are eight letters in the word “DAUGHTER” including three vowels (A, U, E) and 5 consonants (D, G, H, T, R) If the vowels are to be together, we consider them as one letter, so the 6 letters now (5 consonants and 1 vowels entity) can be arranged in 6P6 =
6 ! ways.
How many different eight letter words can be formed out of the word daughter so that?
40320 - 4320 = 36000.
How many different words can be formed from the word daughter so that ending and beginning?
The total number of words formed from 'DAUGHTER' such that no vowels are together is 14400. ⇒nPr=n! n−r! Where n=total number of things and r=no.
How many different words can be formed from the word daughter so that ending and beginning letters are Consonents?
= 40320`. <br> (ii) In the word 'DAUGHTER'. Vowels = A,U,E and Consonants = D,G,T,H,R <br> `rArr` No.
How many words can you make out of daughter?
189 words can be made from the letters in the word daughter.
Can be formed from the letters of the word daughter?
Solution : The letters of the word daughter are “d,a,u,g,h,t,e,r”.
