Class 11 RD Sharma Solutions – Chapter 16 Permutations – Exercise 16.2 | Set 3

Question 33. How many four digit natural numbers not exceeding 4321 can be formed with the digits 1,2,3,4, if the digits can repeat?

Solution:

Total numbers formed by these digits = 4 x 4 x 4 x 4 = 256 (At all 4 positions, 4 options)

Numbers exceeding 4321 are 

At first position 4 is fixed

Case 1: 3 at second position 

3 at 3rd digit = 4 numbers 

4 at 3rd digit = 4 numbers

2 at 3rd and more than digit 1 at 4th = 3 numbers possible   

Case 2: 4 at second position = 4 x 4 = 16 numbers (since 4 possibilities for 3rd as well as the 4th digits)  

Such numbers = Case 1 + Case 2 = 4 + 4 + 3 + 16 = 27 

Total required such numbers = 256 – 27 = 229                           

Question 34. How many numbers of six digits can be formed from the digits 0,1,3,5,7 and 9 when no digit is repeated? How many of them are divisible by 10?

Solution:

First digit = 5 options (except 0)

Remaining = 5! (since 5 positions, 5 digits)

Total numbers = 5 x 5! = 600

Divisible by 10 = 0 at 6th position, remaining 5 digits and 5 positions – 5! = 120 numbers 

Question 35. If three six faced die each marked with numbers 1 to 6 on six faces, are thrown find the total number of possible outcomes.

Solution:

For each dice, number of possible outcomes = 6 

Three dice = 6 x 6 x 6 = 216 possibilities

Question 36. A coin is tossed three times and the outcomes are recorded. How many possible outcomes are there? How many possible outcomes if the coin is tossed four times? Five times? n times?

Solution:

At each toss, 2 different possible outcomes 

3 times coin tossed = 2 x 2 x 2 = 2³ = 8 

4 times = 2 x 2 x 2 x 2 =  2⁴ = 16 

5 times = 2 x 2 x 2 x 2 x 2 = 2⁵ = 32   

n times = 2ⁿ

Question 37. How many numbers of four digits can be formed with the digits 1,2,3,4,5 if the digits can be repeated in the same number?

Solution:

For each digit = 5 possible numbers

So, since 4 digits = 5 x 5 x 5 x 5 = 625 numbers 

Question 38. How many three-digit numbers can be formed by using the digits 0,1,3,5,7 while each digit may be repeated any number of times?

Solution:

For 1st digit = 4 possible

For 2nd & 3rd = 5 possible 

So, 4 x 5 x 5 = 100 such numbers   

Question 39. How many natural numbers less than 1000 can be formed from the digits 0,1,2,3,4,5 when a digit may be repeated any number of times?

Solution:

Given: Total number = 6

As we know that the natural ten’s than 1000 can be 1, 2 and 3 digit numbers 

So, 0 cannot be the first digit of the 3-digit number.

Now, the hundred’s place can be filled with any of the 5 digits = 5 ways

The ten’s place can be filled with any of the 6 digits = 6 ways

The unit place can be filled with any of the 6 digits = 6 ways

Therefore, the total 3 digit number = 5 x 6 x 6 = 180

The total 2 digit number = 5 x 6 = 30

The total 1 digit number = 5 

So, 180 + 30 +1 = 215 natural numbers.

Question 40. How many five digit telephone numbers can be constructed using the digits 0 to 9. If each number starts with 67 and no digit appears more than once?

Solution:

Need to select digits for 3rd, 4th and 5th positions 

So, 8 options for 3rd position and 7 for 4th position and 6 for 5th one

Total numbers are = 8 x 7 x 6 = 336 

Question 41. Find the number of ways in which 8 distinct toys can be distributed among 5 children.

Solution:

Each toy has 5 options. 

So, 8 toys = 5⁸ = 390625

Question 42. Find the number of ways in which one can post 5 letters in 7 letter boxes.

Solution:

Total number of letters = 5

Total number of letter box = 7

So, each letter has 7 options and 5 total letters = 7⁵ = 16807

Question 43. Three dice are rolled. Find the number of possible outcomes in which at least one dice shows 5.

Solution:

Total number of outcomes = 6 x 6 x 6 = 216 

Outcomes having no 5 = 5 x 5 x 5 = 125

Required outcomes = 216 – 125 = 91

Question 44. Find the total number of ways in which 20 balls can be put into 5 boxes so that first box contains just one ball.

Solution:

First box = 20 options

Other 19 balls have = 4 options = 4¹⁹ arrangements for 4 boxes

Total ways = 20 x 4¹⁹

Question 45. In how many ways can 5 different balls be distributed among three boxes?

Solution:

Total number of balls = 5

Total number of boxes =3

So, each ball has 3 options for the boxes = 3⁵ = 243

Question 46. In how many ways can 7 letters be posted in 4 letter boxes?

Solution:

Total number of letters = 7

Total number of letter box = 4

So, each letter has 4 options and 7 total letters = 4⁷ = 16384

Question 47. In how many ways can 4 prizes be distributed among 5 students, when 

(i) No students gets more than one prize?

(ii) A student may get any number of prizes?

(iii) No student gets all the prizes?

Solution:

(i) So, 1 student does not get a prize 

Choose that student among 5 students 5 different possible ways 

Distribute among remaining students = 4! ways = 24 

Total ways = 5 x 24 = 120

(ii) Every prize has 5 options 

4 prizes = 5 x 5 x 5 x 5 = 625 

(iii) Subtract the case of a student to get all prizes from (ii)    

Selecting a student to give all prize =  5 ways 

= 625 – 5 = 620 

Question 48. There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated. 

Solution:

Each lamp has 2 possibilities = on/off 

Total ways = 2¹⁰ = 1024 

One way, all off = need to subtract it 

1024 – 1 = 1023 ways  

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