NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Miscellaneous Exercise

Miscellaneous Exercise

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?

Ans:  There are 3 vowels i.e. A, U and E and 5 consonants i.e. D, G, H, T and R in the given word.

Therefore, the number of ways of selecting 2 vowels out of 3 = $^{3}{{C}_{2}}$=3

Number of ways of selecting 3 consonants out of 5 = $^{5}{{C}_{3}}=10$

Thus, by multiplication principle, the number of combinations = $3\times 10=30$.

Each of these combinations can be arranged in 5! ways.

Hence, the number of different words = $30\times 5!=3600$.

2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that vowels and consonants occur together?

Ans: There are 5 vowels i.e. A, E, I, O and U and 3 consonants i.e. Q, T and N.

Since, vowels and consonants occur together, both (AEIOU) and (QTN) can be considered as single objects.

Thus, there are 5! Permutations of 5 vowels taken all at a time and 3! permutations of 3 consonants taken all at a time.

Therefore, by multiplication principle, the number of words = $2!\, \times \,5!\, \times \,3!\, = \,1440$.

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) at least 3 girls? (iii) at most 3 girls?

Ans: (i) Out of 9 boys and 4 girls, a committee of 7 has to be formed.

Given: exactly 3 girls should be there in a committee, hence, the number of boys = (7 - 3) = 4 boys only.

Therefore, many ways 

$^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}=\frac{4!}{3!1!}\times \frac{9!}{4!5!}$

= $4\times \frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}$

= 504

(ii) Given: at least 3 girls are required in each committee. This can be done in 2 ways

(a) 3 girls and 4 boys or (b) 4 girls and 3 boys

3 girls and 4 boys can be selected in  $^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}$ ways.

4 girls and 3 boys an be selected in  $^{4}{{C}_{4}}{{\times }^{9}}{{C}_{3}}$ ways.

Thus, the number of ways =

$^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}{{+}^{4}}{{C}_{4}}{{\times }^{9}}{{C}_{3}}$

= 504+84

= 588

(iii) Given: at most 3 girls in every committee. This can be done in 4 ways

(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 $^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}$ ways

2 girls and 5 boys can be selected in $^{4}{{C}_{2}}{{\times }^{9}}{{C}_{5}}$ ways

1 girl and 6 boys can be selected in  $^{4}{{C}_{1}}{{\times }^{9}}{{C}_{6}}$ ways

No girl and 7 boys can be selected in $^{4}{{C}_{0}}{{\times }^{9}}{{C}_{7}}$ ways.

Thus, many ways

$ {{=}^{4}}{{C}_{3}}{{\times }^{9}}{{C}_{4}}{{+}^{4}}{{C}_{2}}{{\times }^{9}}{{C}_{5}}{{+}^{4}}{{C}_{1}}{{\times }^{9}}{{C}_{6}}{{+}^{4}}{{C}_{0}}{{\times }^{9}}{{C}_{7}} $

 $ =\frac{4!}{3!1!}\times \frac{9!}{4!5!}+\frac{4!}{2!2!}\times \frac{9!}{5!4!}+\frac{4!}{1!3!}\times \frac{9!}{6!3!}+\frac{4!}{0!4!}\times \frac{9!}{7!2!} $ 

 $ =504+756+336+36 $ 

$=1632 $

4. If the different permutations of all the letters of the word EXAMINATION are listed as in a dictionary, how many words are there in the list before the first word starting with E?

Ans: There are a total of 11 letters out of which A, I and N appear 2 times and other letters appear only once.

The words starting with A will be the words listed before the words starting with E.

Thus, words starting with the letter A will have the letter A fixed at its extreme left end.

And remaining 10 letters are rearranged all at a time.

In the remaining 10 letters, there are 2 I’s and 2 N’s.

Number of words starting with A = $\dfrac{{10!}}{{2!2!}}\, = \,907200$

Therefore, the required number of words is 907200.

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?

Ans: A number can be divisible by 10 only if its unit digit is 0.

Hence, 0 is fixed at the unit place.

Therefore, the 5 vacant places can be filled by the remaining 5 digits.

These 5 empty places can be filled in 5! ways.

Therefore, number of 6-digit numbers = 5! = 120.

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?

Ans: Given: 2 vowels and 2 consonants should be selected from the English alphabet.

We know that there are 5 vowels.

Thus, the number of ways of selecting 2 vowels out of 5 = \[^{5}{{C}_{2}}=\frac{5!}{2!3!}=10\]

Also, we know that there are 21 consonants.

Thus, the number of ways of selecting 2 consonants out of 21 = \[^{21}{{C}_{2}}=\frac{21!}{2!19!}=210\]

Thus, the total number of combinations of selecting 2 vowels and 2 consonants is

= $10\, \times \,210\, = \,2100$.

Every 2100 combination consists of 4 letters, which can be arranged in 4! ways.

Therefore, number of words = $2100\, \times \,4!\, = \,50400$

7. In an examination, a paper consists of 12 Questions divided into 2 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:  Given: 12 Qs are divided into 2 parts – Part I and Part II consisting of 5 and 7 Qs respectively. A student must attempt 8 Qs with at least 3 from each part. This can be done as:

1. 3 Qs from part I and 5 Qs from part II.

2. 4 Qs from part I and 4 Qs from part II.

3. 5 Qs from part I and 3 Qs from part II.

The first case can be selected in $^{5}{{C}_{3}}{{\times }^{7}}{{C}_{5}}$ ways.

The second case can be selected in $^{5}{{C}_{4}}{{\times }^{7}}{{C}_{4}}$ ways.

The third case can be selected in $^{5}{{C}_{5}}{{\times }^{7}}{{C}_{3}}$ ways.

Thus, number of ways of selecting Q’s

${{=}^{5}}{{C}_{3}}{{\times }^{7}}{{C}_{5}}{{+}^{5}}{{C}_{4}}{{\times }^{7}}{{C}_{4}}{{+}^{5}}{{C}_{5}}{{\times }^{7}}{{C}_{3}}$ 

 $ =\frac{5!}{2!3!}\times \frac{7!}{2!5!}+\frac{5!}{4!1!}\times \frac{7!}{4!3!}+\frac{5!}{5!0!}\times \frac{7!}{3!4!} $

$ =210+175+35 $

$=420 $

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.

Ans:  Out 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. We know that in 52 cards, there are 4 kings.

Out of 4 kings, 1 king can be selected in $^{4}{{C}_{1}}$ ways.

Out of the remaining 48 cards, 4 cards can be selected in $^{48}{{C}_{4}}$ ways.

Therefore, number of 5-card combinations =  $^{4}{{C}_{1}}{{\times }^{48}}{{C}_{4}}$ ways.

9. It is required to seat 5 men and 4 women in a row so that the women can occupy the even places. 

How many such arrangements are possible?

Ans: Given 5 men and 4 women should be seated so that women always occupy the even places.

Thus, the men can be seated in 5! ways.

For each arrangement, the women can be seated only in even places.

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

Therefore, possible number of arrangements = $4!\, \times \,5!\, = \,24\, \times \,120\, = \,2880$

10. From the 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:  Given: 10 are chosen for an excursion party out of 25 students.

There are 2 cases where 3 students decided either all or one of them would join.

Case 1: All the 3 students join.

The remaining 7 students can be chosen out of 22 students in $^{22}{{C}_{7}}$ ways.

Case 2: None of the 3 students join.

Thus, 10 students can be chosen in $^{22}{{C}_{10}}$ ways.

Hence, number of ways of choosing excursion party = $^{22}{{C}_{7}}{{+}^{22}}{{C}_{10}}$

11. In how many words can the letters of the word ASSASSINATION be arranged so that all the S’s are together?

Ans: There are 3 A’s, 4 S’s, 2 I’s and all other letters appear only once in the word ASSASSINATION.

The given word should be arranged such that all the S’s are together.

The 4 S’s can be treated as a single object for the time being. This single object with the remaining objects will be 10 objects together.

These 10 objects with 3 A’s, 2 I’s and 2 N’s can be arranged in $\dfrac{{10!}}{{2!3!2!}}$ways.

Therefore, number of ways of arranging the given word = $\dfrac{{10!}}{{2!3!2!}}\, = \,15120$.

Conclusion

Miscellaneous Exercise in Class 11 Maths Chapter 6 is crucial for understanding various concepts thoroughly. It covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorising solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to master this exercise effectively.

Class 11 Maths Chapter 6: Exercises Breakdown

CBSE Class 11 Maths Chapter 6 Other Study Materials

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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