# 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 Ã— 6 Ã— 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 Ã— 9 Ã— 8 Ã— 7

= 80 Ã— 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 Ã—7 Ã— 6 = 56 Ã— 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! â€“ 3!**

**Solution:**

(i) 8!

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

= 40320

(ii) 4! â€“ 3!

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

3! = 3 x 2 x 1 = 6

Therefore, 4! â€“ 3! = 24 â€“ 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 Ã— 5 Ã— 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 = 8C

_{2}

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-1}P_{3}:^{n}P_{4}= 1: 9.**Solution:**

P(n-1, 3) =

P(n, 4) =

=

âˆ´ n = 9

# Question-18

**Find r if : (i) 5P**

_{r}= 2 6P_{r-1}(ii) 5P_{r}= 6P_{r-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 = r^{2}-13r+42

r^{2}-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 =

r^{2 }- 13r + 42 = 6

r^{2 }- 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 â€˜EQUATIONâ€™, using each letter exactly once.**

**Solution:**

There are eight letters in the word â€˜EQUATIONâ€™. 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

_{8}P

_{8}= 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.

(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 iâ€™ s not come together?**

**Solution:**

There are 4Iâ€™s , 1M, 4â€™s and 2Pâ€™s

Word is =

= 110 Ã— 9 Ã— 35

= 990 Ã— 35

= 34650

If the Iâ€™s occur together,

Then the number of arrangements are

= = 35 Ã— 24 = 840

Hence the number of distinct permutations where the Iâ€™s are not together is = 34650â€“ 840 = 33810.

# Question-22

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

**Solution:**

(i) â€˜PERMUTATIONSâ€™ 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**

^{n}C_{8}=^{n}C_{2}, find^{n}C_{2},**Solution:**

Given

^{n}C

_{8}=

^{n}C

_{2}

So 8 + 2 = n

or n = 10

[^{n}C_{a} = ^{n}C_{b} â‡’ a + b = n]

Thus^{ n}C_{2} = ^{10}C_{2 }=

= =

= = 45

# Question-24

**Determine n if (i)**

^{2n}C_{3}: nC_{2}= 12:1 (ii) 2nC_{3}: nC_{3}= 11:1.**Solution:**

(i)=

Ã— =12

= 12

2n-1 = 9

2n = 10

n = 5

(ii) 2nC_{3} : nC_{3} = 11:1

=1

= 11

=

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

8n â€“ 4 = 11n â€“ 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

^{21}C

_{2}, which is given by

^{21}C_{2} = = = 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

= 5C_{3} Ã— 4C_{3} =

= 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 6C

^{3}ways.

Similarly, three white balls can be selected out of 6 white balls in 5C^{3} ways. And three black balls can be selected out of 5 blue balls in 5C^{3} ways. The number ways of selecting 9 balls = 6C^{3} Ã— 5C^{3} Ã— 5C^{3}

=

# 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 4C

_{1}Ã— 48C

_{4}

=

= 4Ã—

= 47 Ã— 46 Ã— 45 Ã— 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 = 5C_{4} Ã— 12C_{7 }

=

=

= 990 Ã— 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

^{5}C

_{2}ways.Similarly, 3 red balls can be selected out of 6 red balls in

^{6}C

_{3}ways.Therefore the number of ways in which 2 black balls and 3 red balls can be drawn is

^{5}C

_{2}Ã—

^{6}C

_{3}.

Now

^{5}C

_{2}Ã—

^{6}C

_{13}= = 10 Ã— 20 = 200

Thus, out of 5 black and 6 red balls, 2 black and 3 red balls can be selected in 200 ways.