# 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

Number of places where the digit can fill

If repetition of digit not allowed,

Then number of Permutations

.

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 thousandâ€™s place = There are 9 ways, as 0 cannot be at thousandâ€™s place

At hundredthâ€™s place = There are 9 digits to be filled, as 1 digit is already taken = 9 ways

At tenâ€™s place = There are now 8 digits to be filled, as 2 digits are already taken = 8 ways

At unitâ€™s place = There are 7 digits that can be filled = 7 ways

The total number of ways to fill the four places = 9 Ã— 9 Ã— 8 Ã— 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 oneâ€™s place = 3 (2, 4 and 6)

Number of permutations

One of the digits is taken at oneâ€™s place; the number of possible digits available = 5

Number of permutations

Therefore, the total number of permutations =3 Ã— 20=60

### Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  OR

3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, = 3 Ã— 5 Ã— 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 oneâ€™s place = 3 (2 and 4 )

Number of permutations

One of the digits is taken at oneâ€™s place; the number of possible digits available = 4

Number of permutations

Therefore, the total number of permutations =2 Ã— 24=48

### Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â 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 Ã— 3 Ã— 2 Ã— 2 = 48

Question 5:Â 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

Number of positions to be filled

Number of permutations

### Â  Â  Â  Â  Â  Â  Â  Â  Â OR

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

Question 6:Â Find n ifÂ n-1P3:Â nP3Â = 1: 9.

Solution: Given that

Question 7: Find r if

(i)Â  5PrÂ = 26Pr-1Â

(ii)Â  5PrÂ =Â 6Pr-1

Solution: (i) Given that

5PrÂ = 26Pr-1

or

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

Therefore

(ii) `Given that,

Â 5PrÂ =Â 6Pr-1

or

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

Therefore

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

= 8 Ã— 7 Ã— 6 Ã— 5 Ã— 4 Ã— 3Ã— 2Ã—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

(ii) Number of letters to be used = 6

Number of permutations

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

â‡’ Number of permutations in vowel

Now, the remaining places = 5

Remaining letters to be used =5

Number of permutations

Therefore, the total number of permutations = 2 Ã— 120 =240

### Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â OR

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

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

Solution: Total number of letters in MISSISSIPPI = 11

Letter number of occurrence

M â†’ 1

I â†’ 4

S â†’ 4

P â†’ 2

Number of permutations

We take that 4 Iâ€™s come together, and they are treated as 1 letter.

âˆ´Â Total number of letters=11 â€“ 4 + 1 = 8

â‡’ Number of permutations

Therefore, the total number of permutations where four Iâ€™s donâ€™t come together = 34650-840=33810

Question 11:Â In how many ways can the letters of the word PERMUTATIONS be arranged if the

(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 â€“ 2 = 10

â‡’ Number of permutationsÂ

(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 â€“ 5 + 1= 8

â‡’ Number of permutations

Therefore, total number of permutations = 120 Ã— 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 Ã— 7 =14

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

âˆ´ No. of permutations

Therefore, the total number of permutations = 14 Ã— 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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