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Question-1

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming (i) repetition of digits allowed. (ii) repetition of digits now allowed?

Solution:
Number of digits given = 5

(i) Since repetition is allowed,

units place can be filled in 5 ways

similarly, tens place can be filled in 5 ways

hundreds place can be filled in 5 ways

Therefore, no. of 3 digit numbers when repetition allowed = 5 x 5 x 5 = 125

(ii) Since repetition is not allowed,

units place can be filled in 5 ways

tens place can be filled in 4 ways

hundreds place can be filled in 3 ways

Therefore, no. of 3 digit numbers when repetition is not allowed = 5 x 4 x 3 = 60

Question-2

How many 3 digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated ?

Solution:
Even numbers are those numbers which has in its unit place the digits 2, 4 and 6.
Therefore the unit digit can be filled up in 3 ways.

However since repetition is allowed, the tens and hundreds place can be filled 6 ways each. Hence, the required number of numbers: 6 × 6 × 3 = 108.

Question-3

How may 4-letter code words are possible using the first 10 letter of the English alphabet, if no letter can be repeated?

Solution:
The number of 4 letter code words out of the first 10 letters of the English

 Alphabets are = 10 × 9 × 8 × 7
                     = 80
× 63
                     = 5040 ways.
 

Question-4

How many 5-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:
Number of 5 digit telephone numbers beginning with 67, with no digit repeated is equivalent to filling 67 - - - with digits.
0, 1, 2, 3, 4, 5, 8, 9, i.e. 8 digits = 8
×7 × 6 = 56 × 6 = 336 ways.

Question-5

A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Solution:
Each time a coin is tossed there are two possible outcomes i.e. head/tail.

Therefore, the possible outcomes when the coin is tossed 3 times = 2 x 2 x 2 = 8

Question-6

Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Solution:
The upper portion can be fitted with 5 colours and the lower portion can be fitted with 4 colours. Therefore, the total number of signals that can be generated using two flags one below the other = 5 x 4 = 20 signals.

Question-7

Evaluate (i) 8! (ii) 4! – 3!

Solution:
(i) 8!

= 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

= 40320

(ii) 4! – 3!

4! = 4 x 3 x 2 x 1 = 24

3! = 3 x 2 x 1 = 6

Therefore, 4! – 3! = 24 – 6 = 18.

Question-8

Is 3! + 4! = 7! ?

Solution:
3! = 6, 4 ! = 24.

Therefore 3! + 4! = 24 + 6 = 30

7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040.

Therefore 3! + 4! Is not equal to 7!.

Question-9

Compute

Solution:
Since,

8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

6!=6 x 5 x 4 x 3 x 2 x 1 and

2!=2 x 1

Thus,

= 4 = 28

Question-10

If find x.

Solution:

=

Therefore x = 64.

Question-11

Evaluate when (i) n = 6, r = 2 (ii) n = 9, r = 5

Solution:
(i)

=

           =

           = 30.

(ii)

          =

          =

          =

          = 15120

Question-12

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

Solution:
1,2,3,4,5,6,7,8,9.

There are 9 digits from 1 to 9.

In the 3 digit number, unit digit can be filled in 9 ways

Ten digit can be filled in 8 ways

Hundredth digit can be filled in 7 ways

Therefore total number of ways = 9 x 8 x 7 = 504 ways.

Question-13

How many four digit numbers are there, with no digit repeated?

Solution:

The digits of the four digit number can be formed using the digits 0, 1, 2, 3, ...9 (Totally 10 digits including '0').

But since a number cannot start with '0' the 1000th digit can be filled using any digit from 1 to 9 (i.e. 9 digits). Hence it can be filled in 9 ways. 

And the 100th place can be filled using any digit from 0 to 9(i.e. 10 digits). Since already one digit is used in the 1000th place the only 9 digits can be used.  Hence it can also be filled in 9 ways. 

Similarly the 10th place and the unit place can be filled in 8 and 7 ways respectively. 

Therefore the total number of ways = 9 x 9 x 8 x 7 = 4536.

Therefore there are 4,536 number of four digit numbers which are not repeated are there.

Question-14

How many even numbers of three digits each can be made with the digits 1, 2, 3, 4, 6, 7 if no digit is repeated?

Solution:
1, 2, 3, 4, 6, 7 -

The units place of the even 3 digit number can be filled up by 2, 4, 6 i.e. 3 numbers.

The tens place by 5 numbers and hundredths place by 4 numbers.

Hence the total of all 3 digit even numbers = 3 × 5 × 4 = 60.

Question-15

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

Solution:
Unit place can be filled using any one of the digits 1,2,3,4 and 5 in 5 ways.
Tenth place can be filled in 4 ways
Hundredth place can be filled in 3 ways
Thousandth place can be filled in 2 ways
Therefore total number of ways = 5 x 4 x 3 x 2 = 120 numbers.
Number of even numbers = Unit digit can be filled in only 2 ways.
And the other, place can be filled in the way as specified above.
Therefore total number of even numbers will be = 2 x 4 x 3 x 2 = 48 numbers.

Question-16

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:
To choose a chairman and a vice chairman from 8 persons = 8C2  
The number of ways in choosing a chairman and a vice chairman such that one person cannot hold more than one position = = = 28.

Question-17

Find n if n-1P3 : nP4 = 1: 9.

Solution:
P(n-1, 3) = 

P(n, 4) =

=



n = 9

Question-18

Find r if :  (i) 5Pr = 2 6Pr-1   (ii) 5Pr = 6Pr-1

Solution:
(i) P(5,r) =

2P(6,r-1) = 2

Given P(5,r) = 2P(6,r-1)

= 2

= 2 dividing by 5!

=

= 1

12 = r2-13r+42

r2-13r-30 = 0

(r - 10)(r - 3) = 0 , r = 10 or 3

Given P(5, r), and since 10 >5, r = 10 is not possible.

Thus, r = 3

(ii) P(5,r) =

P(6,r-1) =

Given =  

=

=

1 =

r2 - 13r + 42 = 6

r2 - 13r + 36 = 0

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

r = 9 or 4

Given P(5, r) and P(6, r-1) r=9 not possible as r >5, thus  r=4.

Question-19

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

Solution:
There are eight letters in the word EQUATION’. So, the total number of words is equal to the number of arrangements of these letters, taken all at a time. The number of such arrangements is 8P8 = 8!. Hence the total number of words = 8!

= 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320 words.

Question-20

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 first letter is a vowel.

Solution:
(i) 4 letters are used at a time:

The number of words can be formed =

=

= 360 words.

(ii) All letters are used at a time:

The number of words that can be formed = 6! = 720 words.

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

The first letter of the word will have the letters O or A i.e. 2 ways.

The remaining places will be 5! Ways.

Therefore total number of words = 5! X 2 = 240 ways.

Question-21

In how many of the distinct permutations of the letters in MISSISSIPPI do the four i’ s not come together?

Solution:
There are 4I’s , 1M, 4’s and 2P’s

Word is =

                     = 110 × 9 × 35

                     = 990 × 35

                     = 34650

If the I’s occur together,

Then the number of arrangements are

= = 35 × 24 = 840

Hence the number of distinct permutations where the I’s are not together is = 34650– 840 = 33810.

Question-22

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

 


Solution:

(i) ‘PERMUTATIONS’ is a 12-letter word.

If the first letter and last letter are fixed as P and S the remaining 10 places will be filled in following way:

The number of repeated letters is: T repeated 2 times.

Therefore possible number of ways is: =

= 1814400

(ii) Vowels are all together

When the vowels are together, these 5 letters will form a place in the word together.

Therefore the number of words will be =
==2419200

Question-23

If nC8 = nC2, find nC2,

Solution:
Given nC8 = nC2

So 8 + 2 = n

or n = 10

[nCa = nCb a + b = n]


Thus nC2 = 10C2 =

= =

= = 45

Question-24

Determine n if   (i) 2nC3: nC2 = 12:1   (ii) 2nC3 : nC3 = 11:1.

Solution:
(i)=

× =12

= 12

2n-1 = 9

2n = 10

n = 5

(ii) 2nC3 : nC3 = 11:1

=1

= 11

=

4(2n - 1) = 11(n - 2)

8n – 4 = 11n – 22

18 = 3n

n = 6

Question-25

How many chords can be drawn through 21 points on a circle?

Solution:
A chord can be drawn by joining any two points on a circle. Therefore, the number of chords that can be drawn out of 21 points is 21C2, which is given by

21C2 = = = 210

Thus, the number of chords is 210.

Question-26

In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Solution:
Number of ways of selecting 3 boys and 3 girls from 5 boys and 4 girls

= 5C3 × 4C3 =

          = 4 = 40

Question-27

Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Solution:
Three red balls can be selected out of 6 red balls in 6C3 ways.

Similarly, three white balls can be selected out of 6 white balls in 5C3 ways. And three black balls can be selected out of 5 blue balls in 5C3 ways. The number ways of selecting 9 balls = 6C3 × 5C3 × 5C3

=

Question-28

Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Solution:
Number of 5 cards combinations from 52 cards with one ace in each
combination is 4C1
× 48C4      
  =       
  
  = 4
×             

  = 47
× 46 × 45 × 8

  = 778320

Question-29

In how many ways can we select a cricket of eleven from 17 players in which only 5 players can bowl if each cricket eleven must include exactly 4 bowlers?

Solution:
The number of ways of selecting a cricket eleven from 17 with 4 bowlers

from 5 players = 5C4 × 12C7

                     =

                     =

                     = 990 × 4

                     = 3960

Question-30

A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Solution:
Two black balls can be selected out of 5 black balls in 5C2 ways.Similarly, 3 red balls can be selected out of 6 red balls in 6C3 ways.Therefore the number of ways in which 2 black balls and 3 red balls can be drawn is 5C2 × 6C3
Now 5C2 × 6C13 = = 10 × 20 = 200
Thus, out of 5 black and 6 red balls, 2 black and 3 red balls can be selected in 200 ways.




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