Class 11 Maths Ncert Solutions Ex 7.3

Class 11 Maths Ncert Solutions Chapter 7 Ex 7.3

Q-1: If there is no repetition of numbers, then find the total of 3- digit numbers formed by using digits from 2 to 9.

 

Solution:

Digits in between 2 to 9 are 8.

We have to construct 3- digit numbers by using the digits from 2 to 9.

It is given that; the repetition of digit is restricted.

So, from the permutation point, there are permutations of 8 digits ( different ) among which we need to take 3 at a time.

Hence, the total of 3- digit number formed by using digits from 2 to 9 = 8P3=8!(83)!=8!5!

= 8×7×6×5!5! = 336

Therefore, total of 336 3-digit numbers can be formed by using digits from 2 to 9 without any repetition.

 

Q-2: Find the total number of 4- digits numbers where there is no repetition of digits.

 

Solution:

We have to construct 4- digit numbers from 0 to 9.

Repetition of digit is restricted.

We know that to have a proper number the thousands place must be filled with 9 digits from 1 to 9, because 0 will be considered as negligible at thousands place.

Therefore, the total number of ways through which the thousands place should be filled is 9.

Now, the hundreds place must be filled with remaining 9 digits between 0 to 9, except that number which is inserted at thousands place, as the repetition is restricted. Similarly, tens and ones place can be filled by the remaining 9 digits.

Hence, for hundreds, tens and ones, there will be 3- digit number and therefore, the permutations of 9 different digits will be taken 3 at the same time.

Now,

The total number of 3- digit number:

= 9P3=9!(93)!=9!6!

= 9×8×7×6!6! = 9 × 8 × 7 = 504

Therefore, by the principle of the multiplication, the total number of the 4 – digit number is 504 × 9 = 4536

 

 

Q-3: Find the total number of 3- digit number which is even and is formed by using the digits 2, 3, 4, 5, 6, 7 and 8. Note that repetition of digit is restricted.

 

Solution:

We need to form even numbers of 3- digits which can be formed by using the digits such as 2, 3, 4, 5, 6, 7 and 8. Repetition of the digit is restricted.

To make the 3- digit number even, the unit place must be filled with an even number that is 2, 4, 6, and 8 among the given number.

As it is given in the question that the repetition of the numbers is restricted so, the unit place must be already occupied with an even number, and the hundreds and tens place is to be filled with the remaining 6 digits given in the question.

Hence, to fill up the hundred and tens place we need to pick up digits except that which is being filled on ones place.  So, we need to pick with permutation of 6 different digits among which 2 digits are taken at a time.

Hence, the number of ways through which the hundreds and tens place is filled:

= 6P2=6!(62)!=9!4!

= 6×5×4!4!  = 30

Therefore, by the principle of the multiplication, the number of the 3- digit numbers required is 4 × 30 = 120.

 

 

Q-4: How many 4- digit numbers will be formed by using the digits 2, 3, 4, 5 and 6. Note that, repetition of digits is not allowed. Find among those numbers, how many numbers will be even number?

 

Solution:

We need to construct 4 – digit numbers using the digits 2, 3, 4, 5 and 6.

We have 5 digits among which we need to form 4- digit number as there are permutations of 5 different digits taken 4 at a time.

Hence, the required number of the 4- digit numbers:

=5P4=5!(54)!=5!1!

= 5×4×3×2×11  = 120

Among these 120 numbers of 4 – digit which is formed by using 2, 3, 4, 5 and 6, even numbers must end with an even number which is 2, 4, or 6.

There are chances that there are 3! Number of ways through which the ones place is filled with digits, i.e., it can be filled by 6 number of ways.

As, in the question, it is mentioned that repetition of digits are restricted and the units place is already occupied with an even number, so the remaining place is to be filled by the remaining 4 digits which is not inserted at any place.

Hence, the number of ways by which the thousands, hundreds and tens place will be filled is the permutation of 4 different digits which is taken 3 at a time.

Therefore, the number of ways to fill the remaining places:

= 4P3=4!(43)!=4!1!

= 4×3×2×11= 24

Hence, by the principle of multiplication, the required number which is an even number is 24 × 3 = 72.

 

 

Q-5: Consider a committee with 9 persons, in how many ways can we choose a chairman and a vice chairman? Note that, a person can hold only one position at a time.

 

Solution:

There is a committee with 9 persons among which, a vice- chairman and a chairman is chosen in such a way that one position can be held by only one person.

So,

To choose a chairman and a vice- chairman, there are number of ways with permutation of 8 different persons taken 2 at a time.

Therefore, the number of ways required:

= 9P2=9!(92)!=9!7!

= 9×8×7!7!= 72

 

 

Q-6: Find n, if n1P3:nP4 = 1 : 9.

 

Solution:

n1P3:nP4=1:9

   n1P3nP4=19

   [(n1)!(n13)!][n!(n4)!]=19

   (n1)!(n4)!×(n4)!n!=19

   (n1)!n×(n1)!=19

1n=19

Therefore, n = 9

Q-7: Find the value of a, if:

(i). 5Pa=2[6Pa1]             (ii). 5Pa=6Pa1

 

Solution:

(i). 5Pa=2[6Pa1]

   5!(5a)!=2×6!(6a+1)!

   5!(5a)!=2×6!(7a)!

   5!(5a)!=2×6×5!(7a)(6a)(5a)!

   1=2×6(7a)(6a)

So, ( 7 – a )  ( 6 – a ) = 12                    

   426a7a+a2=12

   a213a+42=12

   a213a+30=0

   a23a10a+30=0

   a(a3)10(a3)=0

   (a3)(a10)=0

So, a = 3 or a = 10

It is known that, nPr=n!(nr)!,where0an

0a5

Therefore, a ≠ 10

Hence, r = 3

(ii). 5Pa=6Pa1

5!(5a)!=6!(6a+1)!

   5!(5a)!=6!(7a)!

   5!(5a)!=6×5!(7a)(6a)(5a)!

   1=6(7a)(6a)

So, ( 7 – a ) ( 6 – a ) = 6

   426a7a+a2=6

   a213a+42=6

   a213a+36=0

   a24a9a+36=0

   a(a4)9(a4)=0

   (a4)(a9)=0

So, a = 4 or a = 9

It is known that, nPr=n!(nr)!,where0an

0a5

Therefore, a ≠ 9

Hence, a = 4

 

 

Q-8: Get all the number of ways, with or without the meaning, which might be formed by using the letters from the word EQUATION. Note that, repetition of the word is restricted.

 

Solution:

In the word EQUATION, we have 8 letters from the English alphabets.

Hence, to find the number of words formed by using all those 8 letters of the word EQUATION, without repetition of the alphabets, we have:

Permutation of 8 different letters taken 8 at a time, which is 8P8 = 8!

Hence, the number of words required which can be formed = 8! = 40320.

 

 

Q-9: Find the number of words, with or without the meaning, which might be formed by using the letters from the word MONDAY. Note that, repetition of the word is restricted, if

(i) At a time, 3 letters can be used.

(ii) All the letters can be used at a time.

(iii) First letter is a vowel and all the letters are being used.

 

Solution:

In the word MONDAY, we have 6 letters from the English alphabets.

(i) As per the question demand, we need to use 3 letters at a time.

So, the number of words formed by the letters of the word MONDAY, without the repetition of the letters, is

The permutation of 6 different objects which are taken 3 at a time is given by, 6P3

Hence, the number of words required which can be formed by using the word MONDAY using only 3 letters at a time is given by,

6P3 = 6!(63)!=6!3!=6×5×4×3!3!= 6 × 5 × 4 = 120 words

(ii) In this case, at a time, we can use all the letters of the given word MONDAY.

So, to find the number of words formed, we have:

Permutation of 6 different letters taken any of the 6 at a time, that is: 6P6.

Hence, the number of words required which can be formed by using all the letters of the word MONDAY at a time = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.

(iii) In the given words MONDAY, we have two vowels which is O and A.

As per the given condition, vowels must be filled at the start of the word. Also, the repetition of letter is restricted.

So, to this can be done only in 2! Number of ways.

We have 6 letters and the repetition is restricted, also the rightmost place will be filled by a vowel from the word MONDAY, so the remaining 5 places will be filled by the remaining 5 letters. And, this can be done in 5! Number of ways.

Hence, in this case, the number of words required and being formed is 5! × 2! = 5 × 4 × 3 × 2 × 1 × 2 × 1 = 240.

 

 

Q-10: Find the number of distinct permutations such that 4 I’s does not come together in the word MISSISSIPPI.

Solution:

In the word MISSISSIPPI, M appears for 1 time, I appears for 4 times, P appears for 4 times and S appears for 4 times.

So, the required number of the distinct permutations of the letters in the word MISSISSIPPI is given by:

11!4!×4!×2! = 11×10×9×8×7×6×5×4!4!×4×3×2×1×2×1= 11×10×9×8×7×6×54×3×2×1×2×1= 34650.

As there are 4 I’s in this given number which is MISSISSIPPI, whenever they will occur together, they will be treated as a single object which is IIII. Since, all the 4 I’s are considered as 1 and there are 7 letters left in the given word MISSISSIPPI, so we have total of 8 words in account.

Among these 8 words S appears for 4 times and P appears for 4 times, and M appears just once.

So, this can be arranged in 8!4!×2! ways, i.ie., 8 × 7 × 3 × 5 = 840 ways

So, the number of arrangements where all I’s can occur together = 840

Hence, the total number of the distinct permutations of the letters in the given word MISSISSIPPI, where all the 4 I’s won’t come together = 34650 – 840 = 33810.

 

 

Q-11: Arrange the letters of the word PERMUTATIONS and observe the number of ways of the permutation if,

(i). the new word starts with P and ends with S.

(ii). when the vowels are taken all together.

(iii). there will always be 4 letters between P and S.

 

Solution:

In the given word PERMUTATIONS, all the other letters occurred for once except T which occurred for twice.

(i) As we need to fix P and S at the extreme ends i.e., P at the left side of the word and S on the right side of the word, then we have 10 more letters to fix.

Therefore, number of arrangements required in this case = 10!2! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 = 1814400

 

(ii) In the given word PERMUTATIONS, we have all the 5 vowels, i.e., a, e, i, o, u and each of the vowel is appearing only for once in the given word.

As in this case, they have to occur together, so these vowels are considered as a single object. Since, we have 12 letters in the given word and as all the vowels are counted as one, so we have more 7 letters. So, 8 objects are there in the account.

Now,

All these letters can be arranged in 8!2! ways where there is 2 T’s.

As there are 5 different vowels, then corresponding to these arrangements, vowels will be arranged in 5! Ways.

Hence, by the principle of multiplication, number of arrangements required for this case = 8!2!×5! = 2419200.

(iii) We have to arrange the letters of the word PERMUTATIONS in such a way that always there will be 4 letters in between P and S.

So, it is clear that the position of P and S is fixed and 4 others letters can be inserted between P and S. There are 2 T’s in the remaining 10 letters which can be arranged in 10!2! ways.

Also, 4 letters can be placed in between P and S in 2 × 7 = 14 ways.

Hence, by the principle of multiplication, number of arrangements required in this case = 10!2!×14 = 25401600.