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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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Is my ARMY Ready ?
#LetsKillJEE/22
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Mon 21st SepJEE/CBSE 2021: AREA UNDER THE CURVE L-1
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Faster ways of counting
Permutations and Combinations
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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.
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Q1. Find the 2 - digit number (having different digits), which is divisible by 5.
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Solution
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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.
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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.
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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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JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Fundamental principle of multiplication :
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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.
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Q4. How many different words can be formed by arranging the letters of word KNIFE?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Q5. How many different words can be formed by arranging the letters of word ‘DELHI’ such that word starts and end with vowel?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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 .
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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}
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Q7. How many different words can be formed by arranging the letters of word ‘DELHI’ such that word starts and end with vowel?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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 ?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
So by fundamental principle of multiplication total different numbers are 24.
Solution
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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 ?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
So by fundamental principle of multiplication total different numbers are 48.
Solution
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JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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 ?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
• 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
So by fundamental principle of multiplication total different numbers are 27.
As number is odd there is constraint onlast place that it should be odd
Solution
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Q11. How many different three digit numbers are there which contains at least one ‘7’?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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 ?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Maximum number of Total number of
unsuccessful attempts = wrong combinations
– Total correctcombination
Ans =
Ans =
Totalcombinations
10 ×10 ×10 – 1
= 999
Solution
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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 ?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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 ?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
Q16. How many different four letters words can be formed using letters of DAUGHTER such that word contains ‘G’ ?
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
JEE/CBSE 2021: Permutations & Combinations L-1 Fundamental Principle of Counting
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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