NCERT Solutions for Class 11 Maths Chapter 6 - Permutations and Combinations Exercise 6.3

Exercise 6.3

1. How many $3$-digit numbers can be formed by using the digits $1$ to $9$ if no digit is repeated?

Ans:

We have to form a $3$-digit numbers using digits $1-9$. That means, the order of the digits matter.

Hence, there will be as many $3$-digit numbers as there are permutations of $9$ digits taken $3$ at a time.

So, required number of $3$-digit numbers 

$={{\ }^{9}}{{P}_{3}}$

 $\Rightarrow \frac{9!}{(9-3)!}=\frac{9!}{6!}$

$\Rightarrow \frac{9\times 8\times 7\times 6!}{6!}$

$\Rightarrow 9\times 8\times 7=504$

Hence, there are $504$  $3$-digit numbers formed.

2. How many $4$-digit numbers are there with no digit repeated?

Ans:

The thousands place of the $4$-digit number is to be filled with any of the digits from $1$ to $9$ as the digit $0$ cannot be included. Therefore, the number of ways in which thousands place can be filled is $9$. The hundreds, tens, and units place can be filled by any of the digits from $0$ to $9$. However, the digits cannot be repeated in the $4$-digit numbers and thousands place is already occupied with a digit. The hundreds, tens, and units place is to be filled by the remaining $9$-digits. Therefore, there will be as many such $3$-digit numbers as there are permutations of $9$ different digits taken $3$ at a time.

Number of such $3$-digit numbers $={{\ }^{9}}{{P}_{3}}$

$\Rightarrow \frac{9!}{(9-3)!}=\frac{9!}{6!}$

$\Rightarrow \frac{9\times 8\times 7\times 6!}{6!}$

$\Rightarrow 9\times 8\times 7=504$

Thus, by multiplication principle, the required number of $4$-digit numbers are $9\times 504=4536$.

3. How many $3$-digit even numbers can be made using the digits \[\mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{4},\text{ }\mathbf{6},\text{ }\mathbf{7},\] if no digit is repeated?

Ans:

We have to form $3$-digit even numbers using the given six digits, \[1,\text{ }2,\text{ }3,\text{ }4,\text{ }6,\] ,and $7$, without repeating the digits. Then, units can be filled in $3$ ways by any of the digits, $2$, $4$ or $6$. Since the digits cannot be repeated in the $3$-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining $5$-digits. Therefore, the number of ways in which hundreds and tens place can be filled with the remaining $5$-digits is the permutation of $5$ different digits taken $2$ at a time.

So, the number of ways of filling hundreds and tens place

${{\Rightarrow }^{5}}{{P}_{2}}$

$\Rightarrow \frac{5!}{(5-2)!}=\frac{5!}{3!}$

$\Rightarrow \frac{5\times 4\times 3!}{3!}$

$\Rightarrow 5\times 4=20$

Thus, by multiplication principle, the required number of $3$-digit numbers are $3\times 20=60$.

4. Find the number of $4$-digit numbers that can be formed using the digits \[\mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{4},\text{ }\mathbf{5}\] if no digit is repeated. How many of these will be even?

Ans:

We have to form $4$-digit numbers using the digits, $1,\ 2,\ 3,\ 4,$ and $5$. 

So, there will be as many $4$-digit numbers as there are permutations of $5$ different digits taken $4$ at a time.

That is, required number of $4$-digit numbers

${{\Rightarrow }^{5}}{{P}_{4}}$

$\Rightarrow \frac{5!}{(5-4)!}=\frac{5!}{1!}$

$\Rightarrow 1\times 2\times 3\times 4\times 5$

$\Rightarrow 120$

Among the $4$-digit numbers formed by using the digits, $1,\ 2,\ 3,\ 4,\ 5$ even numbers end with either $2$ or $4$. The number of ways in which units place is filled with digits is $2$. Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining $4$-digits. Therefore, the number of ways in which the remaining places can be filled is the permutation of $4$ different digits taken $3$ at time. 

Number of ways of filling the remaining places

${{\Rightarrow }^{4}}{{P}_{3}}=\frac{4!}{(4-3)!}$

$\Rightarrow 4!$

$\Rightarrow 4\times 3\times 2\times 1=24$

Thus, by multiplication principle, the required number of even numbers is $24\times 2=48$.

5. From a committee of $8$ persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

Ans:

We have to choose a chairman and a vice chairman from a committee of $8$ persons in such a way that one person cannot hold more than one position. Here, the number of ways of choosing a chairman and a vice chairman is the permutation of $8$ different objects taken $2$ at a time. 

Thus, required number of ways

${{\Rightarrow }^{8}}{{P}_{2}}=\frac{8!}{(8-2)!}$

$\Rightarrow \frac{8\times 7\times 6!}{6!}=8\times 7$

$\Rightarrow 56$

6. Find $n$ if $^{n-1}{{P}_{3}}{{:}^{n}}{{P}_{4}}=1:9$.

Ans:

Here,

$^{n-1}{{P}_{3}}{{:}^{n}}{{P}_{4}}=1:9$

$\Rightarrow \frac{^{n-1}{{P}_{3}}}{^{n}{{P}_{4}}}=\frac{1}{9}$

$\Rightarrow \frac{\left[ \frac{(n-1)!}{(n-1-3)!} \right]}{\left[ \frac{n!}{(n-4)!} \right]}=\frac{1}{9}$

$\Rightarrow \frac{(n-1)!}{(n-4)!}\times \frac{(n-4)!}{n!}=\frac{1}{9}$

$\Rightarrow \frac{(n-1)!}{n\times (n-1)!}=\frac{1}{9}$

\[\Rightarrow \frac{1}{n}=\frac{1}{9}\]

Hence, $n=9$.

7. Find $r$ if:

(i). $^{5}{{P}_{r}}={{2}^{6}}{{P}_{r-1}}$

Ans:

$\Rightarrow \frac{5!}{(5-r)!}=2\times \frac{6!}{(6-r+1)!}$

$\Rightarrow \frac{5!}{(5-r)!}=\frac{2\times 6\times 5!}{(7-r)(6-r)(5-r)!}$

$\Rightarrow (7-r)(6-r)=12$

$\Rightarrow 42-6r-7r+{{r}^{2}}=12$

$\Rightarrow r(r-3)-10(r-3)=0$

$\Rightarrow r=3\text{ or }r=10$

It is known that $^{n}{{P}_{r}}=\frac{n!}{(n-r)!}$, where $0\le r\le n$.

$\Rightarrow 0\le r\le 5$

Hence, $r=3$.

(ii). $^{5}{{P}_{r}}{{=}^{6}}{{P}_{r-1}}$

Ans: 

$\Rightarrow \frac{5!}{(5-r)!}=\frac{6!}{(6-r+1)!}$

$\Rightarrow \frac{5!}{(5-r)!}=\frac{6\times 5!}{(7-r)(6-r)(5-r)!}$

$\Rightarrow (7-r)(6-r)=6$

$\Rightarrow 42-6r-7r+{{r}^{2}}=6$

$\Rightarrow r(r-4)-9(r-4)=0$

It is known that $^{n}{{P}_{r}}=\frac{n!}{(n-r)!}$, where $0\le r\le n$.

$\Rightarrow 0\le r\le 5$

Hence, $r=4$.

8. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Ans:

There are $8$ different letters in the word EQUATION. Therefore, the number of words that can be formed using all the letters of the word EQUATION, using each letter exactly once, is the number of permutations of $8$ different objects taken $8$ at a time, which is $^{8}{{P}_{8}}=8!$.

Thus, required number of words that can be formed is

$8!=40320$

9. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if

(i). $4$ letters are used at a time,

Ans:

There are 6 different letters in the word MONDAY.

Number of $4$-letter words that can be formed from the letters of the word MONDAY, without repetition of letters, is the number of permutations of $6$ different objects taken $4$ at a time, which is $^{6}{{P}_{4}}$.

Thus, required number of words that can be formed using $4$ letters at a time are $^{6}{{P}_{4}}=\frac{6!}{(6-4)!}$

\[\Rightarrow \frac{6\times 5\times 4\times 3\times 2!}{2!}=6\times 5\times 4\times 3\]

$\Rightarrow 360$

(ii). all letters are used at a time,

Ans:

Number of words that can be formed by using all the letters of the word MONDAY at a time is the number of permutations of $6$ different objects taken $6$ at a time, which is $^{6}{{P}_{6}}=6!$.

Thus, required number of words that can be formed when all letters are used at a time

$6!=1\times 2\times 3\times 4\times 5\times 6$.

$\Rightarrow 720$

(iii). all letters are used but first letter is a vowel.

Ans:

In the given word, there are $2$ different vowels, which have to occupy the rightmost place of the words formed. This can be done only in $2$ ways. Since the letters cannot be repeated and the rightmost place is already occupied with a letter (which is a vowel), the remaining five places are to be filled by the remaining $5$ letters. This can be done in \[5!\] ways.

Thus, in this case, required number of words that can be formed is \[5!\times 2=120\times 2\]

$\Rightarrow 240$

10. In how many of the distinct permutations of the letters in MISSISSIPP do the four I’s not come together?

Ans: 

In the given word MISSISSIPPI, I appear $4$ times, S appears $4$ times, P appears $2$ times, and M appears just once. Therefore, number of distinct permutations of the letters in the given word

$\Rightarrow \frac{11!}{4!4!2!}$

$\Rightarrow \frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{4!\times 4\times 3\times 2\times 1\times 2\times 1}$

$\Rightarrow \frac{11\times 10\times 9\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 1\times 2\times 1}$

$\Rightarrow 34650$

There are $4$ I’s in the given word. When they occur together, they are treated as single object [IIII] for the time being. This single object together with the remaining $7$ objects will account for $8$ objects.

These $8$ objects in which there are $4$ S’s, and $2$ P’s can be arranged in $\frac{8!}{4!2!}$ ways that means, $840$ ways.

Number of arrangements where all I’s occurred together is in $840$ ways.

Thus, number of distinct permutations of the letters in MISSISSIPPI in which four IS do not come together \[34650-840=33810\].

11. In how many ways can the letters of the word PERMUTATIONS be arranged if the 

(i). Words start with P and end with S,

Ans:

In the word PERMUTATIONS, there are $2$ T’s, and all the other letters appear only once.

If P and S are fixed at the extreme ends (P at the left end and S at the right end), then $10$ letters are left.

Hence, in this case, required number of arrangements

$\frac{10!}{2!}=1814400$.

(ii). Vowels are all together,

Ans:

There are $5$ vowels in the given word, each appearing only once.

Since they have to always occur together, they are treated as a single object for the time being. This single object together with the remaining $7$ objects will account for $8$ objects, in total. 

These $8$ objects in which there are $2$ T’s can be arranged in $\frac{8!}{2!}$ ways.

Corresponding to each of these arrangements, the $5$ different vowels can be arranged in $5!$ ways. Therefore, by multiplication principle, required number of arrangements in this case

$\frac{8!}{2!}\times 5!=2419200$.

(iii). There are always $4$ letters between P and S?

Ans:

The letters have to be arranged in such a way that there are always $4$ letters between P and S. Therefore, in a way, the places of P and S are fixed. The remaining $10$ letters in which there are $2$ T’s can be arranged in $\frac{10!}{2!}$ ways.

Since,Possible places of P & S are 1 & 6, 2 & 7, 3 & 8, 4 & 9, 5 & 10, 6 & 11, 7 & 12. Then there can be $4$ letters between P & S.Also P & S can be interchanged as it won’t effect the number of letters between them. 

So, the letters P and S can be placed such that there are $4$ letters between them in $2\times 7=14$ ways.Therefore, by multiplication principle, required number of arrangements in this case

$\frac{10!}{2!}\times 14=25401600$.

Conclusion

In conclusion, Chapter 6 permutations and combinations Ex 6.3 class 11 maths NCERT solutions contains key concepts, emphasizing the importance of order and selection. Focus on understanding the differences between the two, mastering formulas, and practising diverse problems. These topics not only enhance problem-solving skills but also prepare students for more advanced studies in mathematics. Previous year question papers typically feature 3–4 questions on these topics, highlighting their significance in exams. Consistent practice and a clear grasp of concepts are essential for success in this area. 

Class 11 Maths Chapter 6: Exercises Breakdown

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