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JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting

JEE/CBSE 2021: Permutations &

Combinations L-1 Fundamental Principle of

Counting


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Mon 21st SepJEE/CBSE 2021: AREA UNDER THE CURVE L-1


Tues 22nd SepJEE/CBSE 2021: AREA UNDER THE CURVE L- 2 (Between 2 curves)

JEE/CBSE 2021: Permutations & Combinations L-2 Selection

Wed 23rd SepJEE/CBSE 2021: DIFFERENTIAL EQUATIONS L-1 Introduction

JEE/CBSE 2021: Permutations & Combinations L-3 Arrangement

Thu 24th SepJEE/CBSE 2021: DIFFERENTIAL EQUATIONS L-2 Formation

JEE/CBSE 2021: Permutations & Combinations L-4

Fri 25th SepJEE/CBSE 2021: DIFFERENTIAL EQUATIONS L-2 Variable Separable

JEE/CBSE 2021: Permutations & Combinations L-5

Sat 26th Sep

JEE/CBSE 2021: DIFFERENTIAL EQUATIONS L-4 HOMOGENEOUS Form JEE 2021: Permutations & Combinations

Past Year Qs

Class 12th 3pm Class 11th 4.30 pm


Faster ways of counting

Permutations and Combinations


Count the number of people standing below ?

1 2 3 4

5 6 7 8

9 10 11 12

13 14 115 16

17 18 19 20


Let me give you example :Count the number of people standing below ?

There are other methods of counting

1 2 3 4 Ans : 20

5 6 7 84 people in Each Row

9 10 11 12 5 Rowsi.e. 4×5 = 20 people

13 14 115 16


Addition Rule

If work A can be done in m ways and another work B can beDone in n ways and C is a work which is done only when either A or B is completed, then number of ways of doing the work C is m + n.


Q1. Find the 2 - digit number (having different digits), which is divisible by 5.


Solution


Q2. How many three digit numbers divisible by 5 can be formed using any of the digits from 0 to 9 such that none of the digits can be repeated ?

A

B

D

C

136

124

112

108


Solution

We need to find out how many 3 digit numbers divisible by 5 can be formed from the 10 digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) without repetition.

Since the 3 digit number must be divisible by 5, we can have o or 5 at the units place. We will take these as two cases.

Case 1: Three digit numbers ending with 0 Place o at the units place. There is only i way of doing this.

Since the number o is placed at units place, we have now 9 digits (1,2,3,4,5,6,7,8,9) remaining. Any of these 9 digits can be placed at tens place.


Since the digit o is placed at units place and another one digit is placed at tens place, we have now 8 digits remaining. Any of these 8 digits can be placed at hundreds place.

Total number of 3 digit numbers ending with 0 = 8 x 9 x 1 = 72 . . .(A) Case 2: Three digit numbers ending with 5 Place 5 at the units place. There is only 1 way of doing this.

Since the number 5 is placed at units place, we have now 9 digits (0,1,2,3,4,6,7,8,9) remaining. But, from these remaining digits, o cannot be used at hundreds place. Hence any of 8 digits (1,2,3,4,6,7,8,9) can be placed at hundreds place.


Since the digit 5 is placed at units place and another one digit is placed at hundreds place, we have now 8 digits remaining. Any of these 8 digits can be placed at tens place.

Therefore, total number of 3 digit numbers ending with 5 = 8 x 8 x 1 = 64 . . .(B) Hence, required number of 3 digit numbers = 72 + 64 = 136 (∵ from A and B)

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Q2. How many three digit numbers divisible by 5 can be formed using any of the digits from 0 to 9 such that none of the digits can be repeated ?

A

B

D

C

136

124

112

108


Fundamental principle of multiplication :


Fundamental principle of multiplication :

If an event can occur in ‘m’ different ways.Following which another event can occur in ‘n’ different ways.Following which another event can occur in ‘p’ different ways.

Then the total number of ways of simultaneous occurrence of allthese events in a definite order is m × n × p.


Home

1 a

b

c

College

Let me give you example

2Then the total number different routes from Home to College are 2 × 3 = 6We can also Count :1a 1b1c 2a 2b2c


Q3. There are 3 routes from city A to B and 4 routes from city B to C.Find the number of ways for a person to travel from A to C via city B?


A B C

Number of routes from A to B is 3Number of routes from B to C is 4

Solution

As we have to Travel from A to B and then from B to C oneby one simultaneously So Total number of ways are

3 × 4 = 12


Q4. How many different words can be formed by arranging the letters of word KNIFE?


To make the reqd Words we need to Fill 5 Empty Places .As all the places need to be filled one by one simultaneously

Solution

• 1st Empty Space can be filled in 5 ways. {K,N,I,F,E}• After Filling 1st , 2nd empty space can be filled in 4 ways.• After Filling 2nd , 3rd empty space can be filled in 3 ways• After Filling 3rd , 4th empty space can be filled in 2 ways.• After Filling 4th , 5th empty space can be filled in 1 way.

5 × 4 × 3 × 2 × 1 = 120

So by fundamental principle of multiplication total number of ways are 120


Q5. How many different words can be formed by arranging the letters of word ‘DELHI’ such that word starts and end with vowel?


There are constraints on first and last place.

• 1st Empty Space can be filled in 2 ways. {E,I}• After Filling 1st , last empty space can be filled in 1 way.• After Filling 1st & last , 2nd empty space can be filled in 3 ways{D, L,

H}.

• After that 3rd empty space can be filled in 2 ways.

• After that 4th empty space can be filled in 1 way.

2 × 3 × 2 × 1 × 1 = 12So by fundamental principle of multiplication total different words are 12

Solution


Q6. Find the number of 5 letter words which can be formed byletters of word ‘EQUATION’ which start and end with vowel if each letter is to be used at most once .


There are constraints on first and last place.• 1st Empty Space can be filled in 5 ways. {A, E, I, O, U}• After Filling 1st , last empty space can be filled in 4 ways.• After Filling 1st & last , 2nd empty space can be filled in 6 ways.• After that 3rd empty space can be filled in 5 ways.• After that 4th empty space can be filled in 4 ways.

5 × 6 × 5 × 4 × 4 = 2400

So by fundamental principle of multiplication total different words are 2400

Solution We have 5 vowels{A, E, I, O, U} and 3 consonants{Q, T, N}


Q7. How many different words can be formed by arranging the letters of word ‘DELHI’ such that word starts and end with vowel?


To make the reqd numbers we need to Fill 3 Empty Places • 1st Empty Space can be filled in 5 ways. {1,2,4,5,7}

• After Filling 1st , 2nd empty space can be filled in 4 ways.

• After Filling 2nd , 3rd empty space can be filled in 3 ways.

5 × 4 × 3 = 60

So by fundamental principle of multiplication total different numbers are 60.

Solution


Q8. How many 3 digit even numbers can be formed from the digital, 2, 4, 5, 7 if the repetition of the digits is not allowed ?


To make the reqd numbers we need to Fill 3 Empty Places .So there is constraint on last place.

• For even number, last digit should be even. Last place can befilled in 2 ways {2,4}

• After Filling last ,1st empty place can be filled in 4 ways.

• After Filling 1st , 2nd empty place can be filled in 3 ways.4 × 3 × 2 = 24


Solution


Q9. How many 3 digit numbers can be formed from the digits 0, 1, 4, 5, 7if the repetition of the digits is not allowed ?


As number is a ‘3’ digit number first digit cannot be ‘0’

• 1st Empty Space can be filled in 4 ways. {1,4,5,7}

• After Filling 1st , 2nd empty space can be filled in 4 ways.

• After Filling 2nd , 3rd empty space can be filled in 3 ways.

4 × 4 × 3 = 48


Solution

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Q10. How many 3 digit odd numbers can be formed from the digits 0, 1, 4, 5, 7 if the repetition of the digits is not allowed ?


• Last Space can be filled in 3 ways. {1, 5, 7}

• As number is a ‘3’ digit number first digit cannot be ‘0’, So first

empty space can be filled in 3 ways• After Filling in last & 1st , 2nd empty space can be

filled in 3 ways. 3 × 3 × 3 = 27


As number is odd there is constraint onlast place that it should be odd

Solution


Q11. How many different three digit numbers are there which contains at least one ‘7’?


Solution

Reqd numbers = Total 3 digit numbers –3 digit nos. which

do not contain ‘7’

Reqd numbers=

Total 3 digit numbers = 9 × 10 × 10

= 900

3digit numbers which = 8 × 9 × 9do not contain ‘7’

= 648

900 – 648 = 252


Q12. In how many ways five different letters can be posted in 3 mail boxes If each mail box can contain any number of letters ?


Solution

M1 M2 M3

We have to place 5 letters

Number of ways of posting any letter = 3

3 × 3 × 3 × 3 × 3 = 35 = 243So by fundamental principle of multiplication total different ways are 243.

L1 L2 L3 L4 L5


Q13. A number lock has 3 concentric rings with the digits 0, 1, 2, …. , 9 .What is the maximum number of unsuccessfulAttempts to open the lock?


Maximum number of Total number of

unsuccessful attempts = wrong combinations

– Total correctcombination

Ans =

Ans =

Totalcombinations

10 ×10 ×10 – 1

= 999

Solution


Q14. How many different three digit even numbers can be formed using the digits 0, 1, 2, 3, 4, 5 if each digit is to be used at most once ?


Solution

n(Ending with 0) = 5 × 4 × 1 = 20n(Ending with 2 or 4) = 4 × 4 × 2 = 32

As no.s will either end with ‘0’ or with {2, 4} so we have to add all the number of no.s

Ans = 20 + 32 = 52

0

2, 4


Q15. How many three digit even numbers less than 600 can be formed using the digits 0, 1, 2, 3, 4 & 9 if each digit is to be used at most once ?


n(Ending with 0) = 4 × 4 × 1 = 16

Even numbers will either end with ‘0’ or with ‘2’ or with ‘4’

n(Ending with 2 or 4) = 3 × 4 × 2 = 24

As no.s will either end with ‘0’ or with {2, 4}so we have to add all the number of no.s

Ans = 16 + 24 = 40

0

2, 4First digit cannot be‘0’ or ‘9’

Solution


Q16. How many different four letters words can be formed using letters of DAUGHTER such that word contains ‘G’ ?


Solution Reqd words = Total 4 letter words – 4 letter words which

do not contain ‘G’Total 4 letter words

4 letter words which = 7 × 6 × 5 × 4does not contain ‘G’

Reqd words =

= 8 × 7 × 6 × 5

= 1680

=840

1680 – 840

=840


Thank You

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