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# NCERT Miscellaneous Exercise

**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?

**Answer 1:** In the word DAUGHTER, there are 3 vowels namely, A, U, and E, and 5 consonants namely, D, G, H, T, and R.

Number of ways of selecting 2 vowels out of 3 vowels = ^{3}C_{2} = 3

Number of ways of selecting 3 consonants out of 5 consonants = ^{5}C_{3} = 10

Therefore, number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30

Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways.

Hence, required number of different words = 30 × 5! = 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?

**Answer 2:** In the word EQUATION, there are 5 vowels, namely, A, E, I, O, and U, and 3 consonants, namely, Q, T, and N.

Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. Then, the permutations of these 2 objects taken all at a time are counted.

This number would be ^{2}P_{2} = P!

Corresponding to each of these permutations, there are 5! permutations of the five vowels taken all at a time and 3! permutations of the 3 consonants taken all at a time.

Hence, by multiplication principle, required number of words = 2! × 5! × 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?

**Answer 3:** (i) A committee of 7 has to be formed from 9 boys and 4 girls.

Since exactly 3 girls are to be there in every committee, each committee must consist of (7 – 3) = 4 boys only.

Thus, in this case, required number of ways = ^{4}C_{3} x ^{9}C_{4} =

= 504

(ii) Since at least 3 girls are to be there in every committee, the committee can consist of (a) 3 girls and 4 boys or (b) 4 girls and 3 boys

3 girls and 4 boys can be selected in ^{4}C_{3} x ^{9}C_{4} ways.

4 girls and 3 boys can be selected in ^{4}C_{4} x ^{9}C_{3} ways.

Therefore, in this case, required number of ways = ^{4}C_{3} x ^{9}C_{4} + ^{4}C_{4} x ^{9}C_{3}

= 504 + 84 = 588

(iii) Since atmost 3 girls are to be there in every committee, the committee can consist of

(a) 3 girls and 4 boys (b) 2 girls and 5 boys

(c) 1 girl and 6 boys (d) No girl and 7 boys

3 girls and 4 boys can be selected in ^{4}C_{3} x ^{9}C_{4} ways.

2 girls and 5 boys can be selected in ^{4}C_{2} x ^{9}C_{5} ways.

1 girl and 6 boys can be selected in ^{4}C_{1} x ^{9}C_{6} ways.

No girl and 7 boys can be selected in ^{4}C_{0} x ^{9}C_{7} ways.

Therefore, in this case, required number of ways

= ^{4}C_{3} x ^{9}C_{4} + ^{4}C_{2} x ^{9}C_{5 }+ ^{4}C_{1} x ^{9}C_{6 }+ ^{4}C_{0} x ^{9}C_{7}

=

= 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?

**Answer 4:** In the given word EXAMINATION, there are 11 letters out of which, A, I, and N appear 2 times and all the other letters appear only once.

The words that will be listed before the words starting with E in a dictionary will be the words that start with A only.

Therefore, to get the number of words starting with A, the letter A is fixed at the extreme left position, and then the remaining 10 letters taken all at a time are rearranged.

Since there are 2 Is and 2 Ns in the remaining 10 letters,

Number of words starting with A =

Thus, the required numbers of words is 907200.

**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?

**Answer 5: **A number is divisible by 10 if its units digits is 0.

Therefore, 0 is fixed at the units place.

Therefore, there will be as many ways as there are ways of filling 5 vacant places in succession by the remaining 5 digits (i.e., 1, 3, 5, 7 and 9).

The 5 vacant places can be filled in 5! ways.

Hence, required number of 6-digit numbers = 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?

**Answer 6:** 2 different vowels and 2 different consonants are to be selected from the English alphabet.

Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different

vowels from the alphabet = ^{5}C_{2} =

Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet = ^{5}C_{2} =

Therefore, number of combinations of 2 different vowels and 2 different consonants = 10 × 210 = 2100

Each of these 2100 combinations has 4 letters, which can be arranged among themselves in 4! ways.

Therefore, required number of words = 2100 × 4! = 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?

**Answer 7:** It is given that the question paper consists of 12 questions divided into two parts – Part I and Part II, containing 5 and 7 questions, respectively.

A student has to attempt 8 questions, selecting at least 3 from each part.

This can be done as follows.

(a) 3 questions from part I and 5 questions from part II

(b) 4 questions from part I and 4 questions from part II

(c) 5 questions from part I and 3 questions from part II

3 questions from part I and 5 questions from part II can be selected in ^{5}C_{3} x ^{7}C_{5 }ways.

4 questions from part I and 4 questions from part II can be selected in ^{5}C_{4} x ^{7}C_{4 }ways.

5 questions from part I and 3 questions from part II can be selected in ^{5}C_{5} x ^{7}C_{3 }ways.

Thus, required number of ways of selecting questions

= ^{5}C_{3} x ^{7}C_{5 }+ ^{5}C_{4} x ^{7}C_{4 }+ ^{5}C_{5} x ^{7}C_{3}

= 210 + 175 + 35 = 420

**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.

**Answer 8:** From a deck of 52 cards, 5-card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king.

In a deck of 52 cards, there are 4 kings.

1 king can be selected out of 4 kings in ^{4}C_{1 }ways.

4 cards out of the remaining 48 cards can be selected in ^{48}C_{4} ways.

Thus, the required number of 5-card combinations is ^{4}C_{1} x ^{48}C_{4}.

**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?

**Answer 9:** 4 men and 4 women are to be seated in a row such that the women occupy the even places.

The 5 men can be seated in 5! ways. For each arrangement, the 4 women can be seated only at the cross marked places (so that women occupy the even places).

M x M x M x M x M

Therefore, the women can be seated in 4! ways.

Thus, possible number of arrangements = 4! × 5! = 24 × 120 = 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?

**Answer 10:** From the class of 25 students, 10 are to be chosen for an excursion party. Since there are 3 students who decide that either all of them will join or none of them will join, there are two cases.

Case I: All the three students join.

Then, the remaining 7 students can be chosen from the remaining 22 students in ^{22}C_{7 }ways.

Case II: None of the three students join.

Then, 10 students can be chosen from the remaining 22 students in ^{22}C_{10 }ways.

Thus, required number of ways of choosing the excursion party is ^{22}C_{7 }+ ^{22}C_{10}.

**Question 11:** In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?

**Answer 11:** In the given word ASSASSINATION, the letter A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and all the other letters appear only once. Since all the words have to be arranged in such a way that all the Ss are together, SSSS is treated as a single object for the time being. This single object together with the remaining 9 objects will account for 10 objects.

These 10 objects in which there are 3 As, 2 Is, and 2 Ns can be arranged in ways.

Thus, Required number of ways of arranging the letters of the given word =