Q1: How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Ans: The word DAUGHTER has 3 vowels A, E, and U and 5 consonants D, G, H, T and R.
The three vowels can be chosen in 3C2 as only two vowels are to be chosen.
Similarly, the five consonants can be chosen in 5C3 ways.
∴ The number of choosing 2 vowels and 5 consonants would be 3C2 ×5C3
= 30
∴ The total number of ways of is 30.
Each of these 5 letters can be arranged in 5 ways to form different words = 5P5
Total number of words formed would be = 30 × 120 = 3600
Q2: 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?
Ans: In the word EQUATION, there are 5 vowels (A, E, I, O, U) and 3 consonants (Q, T, N).
The numbers of ways in which 5 vowels can be arranged are 5C5
Similarly, the numbers of ways in which 3 consonants can be arranged are 3P3
There are two ways in which vowels and consonants can appear together.
(AEIOU) (QTN) or (QTN) (AEIOU)
∴ The total number of ways in which vowel and consonant can appear together are 2 × 5C5 × 3C3
∴ 2 × 120 × 6 = 1440
Q3: 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) At least 3 girls?
(iii) At most 3 girls?
Ans: (i) Given exactly 3 girls.
The total numbers of girls are 4.
Out of which, 3 are to be chosen.
∴ The number of ways in which choice would be made = 4C3
Numbers of boys are 9 out of which 4 are to be chosen which is given by 9C4
Total ways of forming the committee with exactly three girls.
= 4C3 × 9C4
(ii) Given at least 3 girls.
There are two possibilities for making a committee choosing at least 3 girls.
There are 3 girls and 4 boys, or there are 4 girls and 3 boys.
Choosing three girls we have done in (i)
Choosing four girls and 3 boys would be done in 4C4 ways.
And choosing 3 boys would be done in 9C3
Total ways = 4C4 ×9C3
The total number of ways of making the committee are
504 + 84 = 588
(iii) Given at most 3 girls
In this case, the numbers of possibilities are
0 girl and 7 boys
1 girl and 6 boys
2 girls and 5 boys
3 girls and 4 boys
Number of ways to choose 0 girl and 7 boys = 4C0 × 9C7
The number of choosing 3 girls and 4 boys has been done in (1)
= 504
The total number of ways in which a committee can have at most 3 girls are = 36 + 336 + 756 + 504 = 1632.
Q4: If the different permutations of all the letters of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starts with E?
Ans: In a dictionary, words are listed alphabetically, so to find the words
Listed before E should start with the letter either A, B, C or D.
But the word EXAMINATION doesn’t have B, C or D.
Hence, the words should start with the letter A
The remaining 10 places are to be filled in by the remaining letters of the word EXAMINATION which are E, X, A, M, 2N, T, 2I, 0
Since the letters are repeating, the formula used would be
Where n is the remaining number of letters, p1 and p2 are the number of times the repeated terms occurs.
The number of words in the list before the word starting with E
= words starting with letter A = 907200
Q5: 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?
Ans: The number is divisible by 10 if the unit place has 0 in it.
The 6-digit number is to be formed out of which unit place is fixed as 0.
The remaining 5 places can be filled by 1, 3, 5, 7 and 9.
Here, n = 5
And the numbers of choice available are 5.
So, the total ways in which the rest of the places can be filled are 5P5
Q6: 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?
Ans: We know that there are 5 vowels and 21 consonants in the English alphabet.
Choosing two vowels out of 5 would be done in 5C2 ways.
Choosing 2 consonants out of 21 can be done in 21C2 ways.
The total number of ways to select 2 vowels and 2 consonants
= 5C2 × 21C2
Each of these four letters can be arranged in four ways 4P4
Total numbers of words that can be formed are
24 × 2100 = 50400
Q7: 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?
Ans: The student can choose 3 questions from part I and 5 from part II
Or
4 questions from part I and 4 from part II
5 questions from part 1 and 3 from part II
Q8: Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Ans: We have a deck of cards that has 4 kings.
The numbers of remaining cards are 52.
Ways of selecting a king from the deck = 4C1
Ways of selecting the remaining 4 cards from 48 cards= 48C4
The total number of selecting the 5 cards having one king always
= 4C1 × 48C4
Q9: It is required to seat 5 men and 4 women in a row so that the women occupy even places. How many such arrangements are possible?
Ans: Given there is a total of 9 people.
Women occupy even places, which means they will be sitting in 2nd, 4th, 6th and 8th place where as men will be sitting in 1st, 3rd, 5th,7th and 9th place.
4 women can sit in four places and ways they can be seated= 4P4
5 men can occupy 5 seats in 5 ways.
The number of ways in which these can be seated = 5P5
The total numbers of sitting arrangements possible are
24 × 120 = 2880
Q10: 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?
Ans: In this question, we get 2 options, which are
(i) Either all 3 will go
Then, the remaining students in the class are: 25 – 3 = 22
The number of students remained to be chosen for party = 7
Number of ways to choose the remaining 22 students = 22C7
(ii) None of them will go
The students going will be 10.
Remaining students eligible for going = 22
The number of ways in which these 10 students can be selected are 22C10
The total number of ways in which students can be chosen is
= 170544 + 646646 = 817190
Q11: In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Ans: In the given word ASSASSINATION, there are 4 ‘S’. Since all the 4 ‘S’ have to be arranged together, let us take them as one unit.
The remaining letters are= 3 ‘A’, 2 ‘I’, 2 ‘N’, T
The number of letters to be arranged is 9 (including 4 ‘S’).
Using the formula
where n is the number of terms and p1, p2 p3 are the number of times the repeating letters repeat themselves.
Here, p1= 3, p2= 2, p3 = 2
Putting the values in formula we get
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