Exercise 7.3 ncert solutions maths class 11

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Exercise 7.3(Permutations and Combinations)

Chapter 7 Exercise 7.3 ncert solutions maths class 11

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

Solution: Total number of digit n=9

Number of places where the digit can fill r = 3

If repetition of digit not allowed,

Then number of Permutations = {}^nP_r

=\frac{9!}{(9-3)!}

=\frac{9\times 8 \times 7 \times 6!}{6!}

= 9 \times 8 \times 7

= 504.

Question 2: How many 4-digit numbers are there with no digit repeated?

Solution: Now, we will have 4 places where 4 digits are to be put.

So, at thousand’s place = There are 9 ways, as 0 cannot be at thousand’s place

At hundredth’s place = There are 9 digits to be filled, as 1 digit is already taken = 9 ways

At ten’s place = There are now 8 digits to be filled, as 2 digits are already taken = 8 ways

At unit’s place = There are 7 digits that can be filled = 7 ways

The total number of ways to fill the four places = 9 × 9 × 8 × 7 = 4536 ways

So, a total of 4536 four-digit numbers can be there with no digits repeated.

Question 3: How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Solution: An even number means that the last digit should be even.

Number of possible digits at one’s place = 3 (2, 4 and 6)

Number of permutations = {}^3P_1 =\frac{3!}{(3-1)!}

= 3

One of the digits is taken at one’s place; the number of possible digits available = 5

Number of permutations = {}^5P_2 =\frac{5!}{(5-2)!}

= \frac{5\times 4 \times 3!}{3!} = 20

Therefore, the total number of permutations =3 × 20=60

                        OR

3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, = 3 × 5 × 4 = 60

Question 4: Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, and 5 if no digit is repeated. How many of these will be even?

Solution: An even number means that the last digit should be even.

Number of possible digits at one’s place = 3 (2 and 4 )

Number of permutations = {}^2P_1 =\frac{2!}{(2-1)!}

= 2

One of the digits is taken at one’s place; the number of possible digits available = 4

Number of permutations = {}^4P_3 =\frac{4!}{(4-3)!}

=\frac{4!}{1!}

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

Therefore, the total number of permutations =2 × 24=48

                                 OR

The number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, and 5 if no digit is repeated = 4 × 3 × 2 × 2 = 48

Question 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?

Solution: Total number of people in committee n = 8

Number of positions to be filled r= 2

Number of permutations = {}^8P_2

= \frac{8!}{(8-2)!}

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

= 8 \times 7=56

                 OR

Number of way of selection of chairman and a vice chairman from 8 person = 8 × 7 = 56

Question 6: Find n if n-1P3: nP3 = 1: 9.

Solution: Given that

\dfrac{{}^{n-1}P_3}{{}^{n}P_3}=\dfrac{1}{9}

\Rightarrow \drac{\frac{(n-1)!}{(n-4)!}}{\frac{n!}{(n-4)!}}=\dfrac{1}{9}

\Rightarrow \dfrac{(n-1)!}{(n)!}=\dfrac{1}{9}

\Rightarrow \dfrac{(n-1)}{n(n-1)!}=\dfrac{1}{9}

\Rightarrow \dfrac{1}{n} = \dfrac{1}{9}

\Rightarrow n = 9

Question 7: Find r if

(i)  5Pr = 26Pr-1 

(ii)  5Pr = 6Pr-1

Solution: (i) Given that

5Pr = 26Pr-1

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

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

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

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

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

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

\Rightarrow r^2 - 13r +42-12=0

\Rightarrow r^2 -13r+30=0

\Rightarrow r^2-10r-3r+30=0

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

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

\Rightarrow r = 10 or r = 3

But r = 10 is rejected, r can not be greater than 5,

Therefore r = 3

(ii) `Given that,

 5Pr = 6Pr-1

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

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

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

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

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

\Rightarrow r^2 - 13r +42-6=0

\Rightarrow r^2 -13r+36=0

\Rightarrow r^2-9r-4r+36=0

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

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

\Rightarrow r = 9 or r = 4

But r = 9 is rejected, r can not be greater than 5,

Therefore r = 4

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

Solution: Total number of different letters in EQUATION = 8

Number of letters to be used to form a word = 8

Number of permutations = {}^8P_8

=\frac{8!}{(8-8)!}

= \frac{8!}{0!}

= 8 × 7 × 6 × 5 × 4 × 3× 2×1

= 40320

Question 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

(ii) All letters are used at a time

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

Solution: (i) Number of letters to be used =4

Number of permutations = {}^6P_4

=\frac{6!}{(6-4)!}

= \frac{6!}{2!}

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

=360

(ii) Number of letters to be used = 6

Number of permutations = {}^6P_6

=\frac{6!}{(6-6)!}

= \frac{6!}{0!}

=6\times 5\times 4\times 3\times 2\times 1

=720

(iii) Number of vowels in MONDAY = 2 (O and A)

⇒ Number of permutations in vowel = {}^2P_1=2

Now, the remaining places = 5

Remaining letters to be used =5

Number of permutations = {}^5P_5

=\frac{5!}{(5-5)!}

= \frac{5!}{0!}

= 5\times 4\times 3\times 2\times 1

=120

Therefore, the total number of permutations = 2 × 120 =240

                                 OR

all letters are used, but the first letter is a vowel = 2×5×4×3×2×1 = 240

Question 10: In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

Solution: Total number of letters in MISSISSIPPI = 11

Letter number of occurrence

M → 1

I → 4

S → 4

P → 2

Number of permutations = \frac{(11)!}{1!4!4!2!}=34650

We take that 4 I’s come together, and they are treated as 1 letter.

∴ Total number of letters=11 – 4 + 1 = 8

⇒ Number of permutations = \frac{8!}{1!4!2!}=840

Therefore, the total number of permutations where four I’s don’t come together = 34650-840=33810

Question 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

(ii) Vowels are all together

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

Solution: (i) Total number of letters in PERMUTATIONS =12

The only repeated letter is T = 2times

The first and last letters of the word are fixed as P and S, respectively.

Number of letters remaining =12 – 2 = 10

⇒ Number of permutations  =\dfrac{{}^{10}P_{10}}{2!}

= \dfrac{10!}{2(10-10)!}

= \dfrac{10!}{2} = 1814400

(ii) Number of vowels in PERMUTATIONS = 5 (E, U, A, I, O)

Now, we consider all the vowels together as one.

Number of permutations of vowels = 120

Now total number of letters = 12 – 5 + 1= 8

⇒ Number of permutations = \dfrac{{}^8P_8}{2}

=\dfrac{8!}{2}

=20160

Therefore, total number of permutations = 120 × 20160 = 2419200

(iii) Number of places is as 1 2 3 4 5 6 7 8 9 10 11 12

There should always be 4 letters between P and S.

Possible places of P and S are 1 and 6, 2 and 7, 3 and 8, 4 and 9, 5 and 10, 6 and 11, 7 and 12

Possible ways =7

Also, P and S can be interchanged.

No. of permutations =2 × 7 =14

The remaining 10 places can be filled with 10 remaining letters,

∴ No. of permutations =\dfrac{{}^{10}P_{10}}{2!}

= \dfrac{10!}{2(10-10)!}

= \dfrac{10!}{2} = 1814400

Therefore, the total number of permutations = 14 × 1814400 =25401600

Chapter 7 Permutation and Combination Ncert math solution

Exercise 7.1 Permutation and Combination Ncert math solution

Exercise 7.2 Permutation and Combination Ncert math solution

Exercise 7.3 Permutation and Combination Ncert math solution

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