Permutations and Combinations: JEE Main Previous Year Questions (2021- 2024) | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


JEE Mains Previous Year Questions 
(2021-2024): Permutations and Combinations 
2024 
Q1: If ?? is the number of ways five different employees can sit into four indistinguishable offices 
where any office may have any number of persons including zero, then ?? is equal to: 
A. 47 
B. 53 
C. 51 
D. 43   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Explanation: Currently no explanation available 
Q2: The number of ways in which 21 identical apples can be distributed among three children such 
that each child gets at least 2 apples, is 
A. 130 
B. 136 
C. 142 
D. 406   [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (b) 
To solve this problem, we can use a classic combinatorics method known as "stars and bars" (or "balls 
and bins"), which is a way to solve problems involving distributing identical items into distinct groups 
with certain restrictions. 
First, since each child must get at least 2 apples, let's give 2 apples to each child right away. That 
accounts for 6 apples ( 2 apples for each of the 3 children). Now, we have 21 - 6 = 15 apples left to 
distribute freely among the three children. 
The "stars and bars" technique involves representing the apples as stars ( 
*
) and the divisions between 
children as bars (I). For example, if we had 5 apples to distribute among three children, one possible 
distribution could be represented as  
**
|
*
|
*?
. This means the first child gets 2 apples, the second child 
gets 1 apple, and the third child gets 2 apples. 
In our case, we need to distribute 15 apples (stars) among the three children with 2 bars to create the 
partitions. We arrange 15 stars and 2 bars in a row, where the arrangement of stars and bars 
corresponds to a distribution of the apples. 
The total number of objects we're arranging is 15 apples +2 bars = 17 objects. We need to choose 2 
positions out of these 17 to place the bars. The remaining positions will be occupied by the stars 
(apples). 
The total number of objects we're arranging is 15 apples +2 bars = 17 objects. We need to choose 2 
Page 2


JEE Mains Previous Year Questions 
(2021-2024): Permutations and Combinations 
2024 
Q1: If ?? is the number of ways five different employees can sit into four indistinguishable offices 
where any office may have any number of persons including zero, then ?? is equal to: 
A. 47 
B. 53 
C. 51 
D. 43   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Explanation: Currently no explanation available 
Q2: The number of ways in which 21 identical apples can be distributed among three children such 
that each child gets at least 2 apples, is 
A. 130 
B. 136 
C. 142 
D. 406   [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (b) 
To solve this problem, we can use a classic combinatorics method known as "stars and bars" (or "balls 
and bins"), which is a way to solve problems involving distributing identical items into distinct groups 
with certain restrictions. 
First, since each child must get at least 2 apples, let's give 2 apples to each child right away. That 
accounts for 6 apples ( 2 apples for each of the 3 children). Now, we have 21 - 6 = 15 apples left to 
distribute freely among the three children. 
The "stars and bars" technique involves representing the apples as stars ( 
*
) and the divisions between 
children as bars (I). For example, if we had 5 apples to distribute among three children, one possible 
distribution could be represented as  
**
|
*
|
*?
. This means the first child gets 2 apples, the second child 
gets 1 apple, and the third child gets 2 apples. 
In our case, we need to distribute 15 apples (stars) among the three children with 2 bars to create the 
partitions. We arrange 15 stars and 2 bars in a row, where the arrangement of stars and bars 
corresponds to a distribution of the apples. 
The total number of objects we're arranging is 15 apples +2 bars = 17 objects. We need to choose 2 
positions out of these 17 to place the bars. The remaining positions will be occupied by the stars 
(apples). 
The total number of objects we're arranging is 15 apples +2 bars = 17 objects. We need to choose 2 
positions out of these 17 to place the bars. The remaining positions will be occupied by the stars 
(apples). 
The number of ways to choose 2 positions out of 17 for the bars is given by the binomial coefficient: 
Number of ways = (
17
2
) =
17!
2!(17-2)!
=
17×16
2×1
= 136 
Thus, there are 136 ways to distribute the 21 identical apples among three children such that each child 
gets at least 2 apples. The correct answer is Option B: 136 . 
Q3: If for some ?? , ?? ; 
?? ?? ?? + ?? ( 
?? ?? ?? +?? ) + 
?? ?? ?? +?? > 
?? ?? ?? and  
?? -?? ?? ?? : 
?? ?? ?? = ?? : ?? , then  
?? ?? ?? +?? +
 
?? +?? ?? ?? is equal to 
A. 380 
B. 376 
C. 372 
D. 384    [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (c) 
 
6
C
m
+ 2( 
6
C
m+1
) + 
6
C
m+2
> 
8
C
3
 
7
C
m+1
+ 
7
C
m+2
> 
8
C
3
 
8
C
m+2
> 
8
C
3
 ? m = 2
 And  
n-1
P
3
: 
?? P
4
= 1: 8
(n - 1)(n - 2)(n - 3)
n(n - 1)(n - 2)(n - 3)
=
1
8
 ? n = 8
 ? 
n
P
m+1
+ 
n+1
C
m
= 
8
P
3
+ 
9
C
2
 = 8 × 7 × 6 +
9 × 8
2
 = 372
 
Q4: Number of ways of arranging 8 identical books into 4 identical shelves where any number of 
shelves may remain empty is equal to 
A. 18 
B. 16 
C. 12 
D. 15    [JEE Main 2024 (Online) 29th January Evening Shift] 
Page 3


JEE Mains Previous Year Questions 
(2021-2024): Permutations and Combinations 
2024 
Q1: If ?? is the number of ways five different employees can sit into four indistinguishable offices 
where any office may have any number of persons including zero, then ?? is equal to: 
A. 47 
B. 53 
C. 51 
D. 43   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Explanation: Currently no explanation available 
Q2: The number of ways in which 21 identical apples can be distributed among three children such 
that each child gets at least 2 apples, is 
A. 130 
B. 136 
C. 142 
D. 406   [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (b) 
To solve this problem, we can use a classic combinatorics method known as "stars and bars" (or "balls 
and bins"), which is a way to solve problems involving distributing identical items into distinct groups 
with certain restrictions. 
First, since each child must get at least 2 apples, let's give 2 apples to each child right away. That 
accounts for 6 apples ( 2 apples for each of the 3 children). Now, we have 21 - 6 = 15 apples left to 
distribute freely among the three children. 
The "stars and bars" technique involves representing the apples as stars ( 
*
) and the divisions between 
children as bars (I). For example, if we had 5 apples to distribute among three children, one possible 
distribution could be represented as  
**
|
*
|
*?
. This means the first child gets 2 apples, the second child 
gets 1 apple, and the third child gets 2 apples. 
In our case, we need to distribute 15 apples (stars) among the three children with 2 bars to create the 
partitions. We arrange 15 stars and 2 bars in a row, where the arrangement of stars and bars 
corresponds to a distribution of the apples. 
The total number of objects we're arranging is 15 apples +2 bars = 17 objects. We need to choose 2 
positions out of these 17 to place the bars. The remaining positions will be occupied by the stars 
(apples). 
The total number of objects we're arranging is 15 apples +2 bars = 17 objects. We need to choose 2 
positions out of these 17 to place the bars. The remaining positions will be occupied by the stars 
(apples). 
The number of ways to choose 2 positions out of 17 for the bars is given by the binomial coefficient: 
Number of ways = (
17
2
) =
17!
2!(17-2)!
=
17×16
2×1
= 136 
Thus, there are 136 ways to distribute the 21 identical apples among three children such that each child 
gets at least 2 apples. The correct answer is Option B: 136 . 
Q3: If for some ?? , ?? ; 
?? ?? ?? + ?? ( 
?? ?? ?? +?? ) + 
?? ?? ?? +?? > 
?? ?? ?? and  
?? -?? ?? ?? : 
?? ?? ?? = ?? : ?? , then  
?? ?? ?? +?? +
 
?? +?? ?? ?? is equal to 
A. 380 
B. 376 
C. 372 
D. 384    [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (c) 
 
6
C
m
+ 2( 
6
C
m+1
) + 
6
C
m+2
> 
8
C
3
 
7
C
m+1
+ 
7
C
m+2
> 
8
C
3
 
8
C
m+2
> 
8
C
3
 ? m = 2
 And  
n-1
P
3
: 
?? P
4
= 1: 8
(n - 1)(n - 2)(n - 3)
n(n - 1)(n - 2)(n - 3)
=
1
8
 ? n = 8
 ? 
n
P
m+1
+ 
n+1
C
m
= 
8
P
3
+ 
9
C
2
 = 8 × 7 × 6 +
9 × 8
2
 = 372
 
Q4: Number of ways of arranging 8 identical books into 4 identical shelves where any number of 
shelves may remain empty is equal to 
A. 18 
B. 16 
C. 12 
D. 15    [JEE Main 2024 (Online) 29th January Evening Shift] 
Ans: (d) 
 
Q5: Let ?? and ?? be two finite sets with ?? and ?? elements respectively. The total number of subsets of 
the set ?? is 56 more than the total number of subsets ?? . Then the distance of the point ?? (?? , ?? ) from 
the point ?? (-?? , -?? ) is : 
A. 8 
B. 10 
C. 4 
D. 6    [JEE Main 2024 (Online) 27th January Evening Shift] 
Ans: (b) 
Explanation: Currently no explanation available 
Q6: The number of elements in the set ?? = {(?? , ?? , ?? ): ?? , ?? , ?? ? ?? , ?? + ?? ?? + ?? ?? = ???? , ?? , ?? , ?? ? ?? } equals 
[JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: 169 
Explanation: Currently no explanation available 
Q7: The total number of words (with or without meaning) that can be formed out of the letters of the 
word 'DISTRIBUTION' taken four at a time, is equal to    [JEE Main 2024 (Online) 31st January Morning 
Shift] 
Ans: 3734 
We have III, TT, D, S, R, B, U, O, N 
Number of words with selection (a, a, a, b) 
= 
8
C
1
×
4!
3!
= 32 
Number of words with selection (a, a, b, b) 
Page 4


JEE Mains Previous Year Questions 
(2021-2024): Permutations and Combinations 
2024 
Q1: If ?? is the number of ways five different employees can sit into four indistinguishable offices 
where any office may have any number of persons including zero, then ?? is equal to: 
A. 47 
B. 53 
C. 51 
D. 43   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Explanation: Currently no explanation available 
Q2: The number of ways in which 21 identical apples can be distributed among three children such 
that each child gets at least 2 apples, is 
A. 130 
B. 136 
C. 142 
D. 406   [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (b) 
To solve this problem, we can use a classic combinatorics method known as "stars and bars" (or "balls 
and bins"), which is a way to solve problems involving distributing identical items into distinct groups 
with certain restrictions. 
First, since each child must get at least 2 apples, let's give 2 apples to each child right away. That 
accounts for 6 apples ( 2 apples for each of the 3 children). Now, we have 21 - 6 = 15 apples left to 
distribute freely among the three children. 
The "stars and bars" technique involves representing the apples as stars ( 
*
) and the divisions between 
children as bars (I). For example, if we had 5 apples to distribute among three children, one possible 
distribution could be represented as  
**
|
*
|
*?
. This means the first child gets 2 apples, the second child 
gets 1 apple, and the third child gets 2 apples. 
In our case, we need to distribute 15 apples (stars) among the three children with 2 bars to create the 
partitions. We arrange 15 stars and 2 bars in a row, where the arrangement of stars and bars 
corresponds to a distribution of the apples. 
The total number of objects we're arranging is 15 apples +2 bars = 17 objects. We need to choose 2 
positions out of these 17 to place the bars. The remaining positions will be occupied by the stars 
(apples). 
The total number of objects we're arranging is 15 apples +2 bars = 17 objects. We need to choose 2 
positions out of these 17 to place the bars. The remaining positions will be occupied by the stars 
(apples). 
The number of ways to choose 2 positions out of 17 for the bars is given by the binomial coefficient: 
Number of ways = (
17
2
) =
17!
2!(17-2)!
=
17×16
2×1
= 136 
Thus, there are 136 ways to distribute the 21 identical apples among three children such that each child 
gets at least 2 apples. The correct answer is Option B: 136 . 
Q3: If for some ?? , ?? ; 
?? ?? ?? + ?? ( 
?? ?? ?? +?? ) + 
?? ?? ?? +?? > 
?? ?? ?? and  
?? -?? ?? ?? : 
?? ?? ?? = ?? : ?? , then  
?? ?? ?? +?? +
 
?? +?? ?? ?? is equal to 
A. 380 
B. 376 
C. 372 
D. 384    [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (c) 
 
6
C
m
+ 2( 
6
C
m+1
) + 
6
C
m+2
> 
8
C
3
 
7
C
m+1
+ 
7
C
m+2
> 
8
C
3
 
8
C
m+2
> 
8
C
3
 ? m = 2
 And  
n-1
P
3
: 
?? P
4
= 1: 8
(n - 1)(n - 2)(n - 3)
n(n - 1)(n - 2)(n - 3)
=
1
8
 ? n = 8
 ? 
n
P
m+1
+ 
n+1
C
m
= 
8
P
3
+ 
9
C
2
 = 8 × 7 × 6 +
9 × 8
2
 = 372
 
Q4: Number of ways of arranging 8 identical books into 4 identical shelves where any number of 
shelves may remain empty is equal to 
A. 18 
B. 16 
C. 12 
D. 15    [JEE Main 2024 (Online) 29th January Evening Shift] 
Ans: (d) 
 
Q5: Let ?? and ?? be two finite sets with ?? and ?? elements respectively. The total number of subsets of 
the set ?? is 56 more than the total number of subsets ?? . Then the distance of the point ?? (?? , ?? ) from 
the point ?? (-?? , -?? ) is : 
A. 8 
B. 10 
C. 4 
D. 6    [JEE Main 2024 (Online) 27th January Evening Shift] 
Ans: (b) 
Explanation: Currently no explanation available 
Q6: The number of elements in the set ?? = {(?? , ?? , ?? ): ?? , ?? , ?? ? ?? , ?? + ?? ?? + ?? ?? = ???? , ?? , ?? , ?? ? ?? } equals 
[JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: 169 
Explanation: Currently no explanation available 
Q7: The total number of words (with or without meaning) that can be formed out of the letters of the 
word 'DISTRIBUTION' taken four at a time, is equal to    [JEE Main 2024 (Online) 31st January Morning 
Shift] 
Ans: 3734 
We have III, TT, D, S, R, B, U, O, N 
Number of words with selection (a, a, a, b) 
= 
8
C
1
×
4!
3!
= 32 
Number of words with selection (a, a, b, b) 
=
4!
2! 2!
= 6 
Number of words with selection (?? , ?? , ?? , ?? ) 
= 
2
C
1
× 
8
C
2
×
4!
2!
= 672 
Number of words with selection (a, b, c, d) 
 = 
9
C
4
× 4! = 3024
 ? total = 3024 + 672 + 6 + 32
 = 3734
 
Q8: In an examination of Mathematics paper, there are 20 questions of equal marks and the question 
paper is divided into three sections : ?? , ?? and ?? . A student is required to attempt total 15 questions 
taking at least 4 questions from each section. If section ?? has 8 questions, section ?? has 6 questions 
and section ?? has 6 questions, then the total number of ways a student can select 15 questions is 
[JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: 11376 
If 4 questions from each section are selected 
Remaining 3 questions can be selected either in (1,1,1) or (3,0,0) or (2,1,0) 
 ? Total ways = 
8
c
5
· 
6
c
5
· 
6
c
5
+ 
8
c
6
 
6
c
5
· 
6
c
4
× 2 +
 
8
c
5
· 
6
c
6
· 
6
c
4
× 2 + 
8
c
4
· 
6
c
6
· 
6
c
5
× 2 + 
8
c
7
· 
6
c
4
· 
6
c
4
 = 56.6.6 + 28.6.15.2 + 56.15.2 + 70.6.2
 +8.15.15
 = 2016 + 5040 + 1680 + 840 + 1800 = 11376
 
Q9: If 
 
????
?? ?? ?? +
 
????
?? ?? ?? + ? +
 
????
?? ?? ????
=
?? ?? with ?????? (?? , ?? ) = ?? , then ?? + ?? is equal to    [JEE Main 2024 
(Online) 29th January Morning Shift] 
Ans: 2041 
Explanation: Currently no explanation available 
 
 
 
 
Page 5


JEE Mains Previous Year Questions 
(2021-2024): Permutations and Combinations 
2024 
Q1: If ?? is the number of ways five different employees can sit into four indistinguishable offices 
where any office may have any number of persons including zero, then ?? is equal to: 
A. 47 
B. 53 
C. 51 
D. 43   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Explanation: Currently no explanation available 
Q2: The number of ways in which 21 identical apples can be distributed among three children such 
that each child gets at least 2 apples, is 
A. 130 
B. 136 
C. 142 
D. 406   [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (b) 
To solve this problem, we can use a classic combinatorics method known as "stars and bars" (or "balls 
and bins"), which is a way to solve problems involving distributing identical items into distinct groups 
with certain restrictions. 
First, since each child must get at least 2 apples, let's give 2 apples to each child right away. That 
accounts for 6 apples ( 2 apples for each of the 3 children). Now, we have 21 - 6 = 15 apples left to 
distribute freely among the three children. 
The "stars and bars" technique involves representing the apples as stars ( 
*
) and the divisions between 
children as bars (I). For example, if we had 5 apples to distribute among three children, one possible 
distribution could be represented as  
**
|
*
|
*?
. This means the first child gets 2 apples, the second child 
gets 1 apple, and the third child gets 2 apples. 
In our case, we need to distribute 15 apples (stars) among the three children with 2 bars to create the 
partitions. We arrange 15 stars and 2 bars in a row, where the arrangement of stars and bars 
corresponds to a distribution of the apples. 
The total number of objects we're arranging is 15 apples +2 bars = 17 objects. We need to choose 2 
positions out of these 17 to place the bars. The remaining positions will be occupied by the stars 
(apples). 
The total number of objects we're arranging is 15 apples +2 bars = 17 objects. We need to choose 2 
positions out of these 17 to place the bars. The remaining positions will be occupied by the stars 
(apples). 
The number of ways to choose 2 positions out of 17 for the bars is given by the binomial coefficient: 
Number of ways = (
17
2
) =
17!
2!(17-2)!
=
17×16
2×1
= 136 
Thus, there are 136 ways to distribute the 21 identical apples among three children such that each child 
gets at least 2 apples. The correct answer is Option B: 136 . 
Q3: If for some ?? , ?? ; 
?? ?? ?? + ?? ( 
?? ?? ?? +?? ) + 
?? ?? ?? +?? > 
?? ?? ?? and  
?? -?? ?? ?? : 
?? ?? ?? = ?? : ?? , then  
?? ?? ?? +?? +
 
?? +?? ?? ?? is equal to 
A. 380 
B. 376 
C. 372 
D. 384    [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (c) 
 
6
C
m
+ 2( 
6
C
m+1
) + 
6
C
m+2
> 
8
C
3
 
7
C
m+1
+ 
7
C
m+2
> 
8
C
3
 
8
C
m+2
> 
8
C
3
 ? m = 2
 And  
n-1
P
3
: 
?? P
4
= 1: 8
(n - 1)(n - 2)(n - 3)
n(n - 1)(n - 2)(n - 3)
=
1
8
 ? n = 8
 ? 
n
P
m+1
+ 
n+1
C
m
= 
8
P
3
+ 
9
C
2
 = 8 × 7 × 6 +
9 × 8
2
 = 372
 
Q4: Number of ways of arranging 8 identical books into 4 identical shelves where any number of 
shelves may remain empty is equal to 
A. 18 
B. 16 
C. 12 
D. 15    [JEE Main 2024 (Online) 29th January Evening Shift] 
Ans: (d) 
 
Q5: Let ?? and ?? be two finite sets with ?? and ?? elements respectively. The total number of subsets of 
the set ?? is 56 more than the total number of subsets ?? . Then the distance of the point ?? (?? , ?? ) from 
the point ?? (-?? , -?? ) is : 
A. 8 
B. 10 
C. 4 
D. 6    [JEE Main 2024 (Online) 27th January Evening Shift] 
Ans: (b) 
Explanation: Currently no explanation available 
Q6: The number of elements in the set ?? = {(?? , ?? , ?? ): ?? , ?? , ?? ? ?? , ?? + ?? ?? + ?? ?? = ???? , ?? , ?? , ?? ? ?? } equals 
[JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: 169 
Explanation: Currently no explanation available 
Q7: The total number of words (with or without meaning) that can be formed out of the letters of the 
word 'DISTRIBUTION' taken four at a time, is equal to    [JEE Main 2024 (Online) 31st January Morning 
Shift] 
Ans: 3734 
We have III, TT, D, S, R, B, U, O, N 
Number of words with selection (a, a, a, b) 
= 
8
C
1
×
4!
3!
= 32 
Number of words with selection (a, a, b, b) 
=
4!
2! 2!
= 6 
Number of words with selection (?? , ?? , ?? , ?? ) 
= 
2
C
1
× 
8
C
2
×
4!
2!
= 672 
Number of words with selection (a, b, c, d) 
 = 
9
C
4
× 4! = 3024
 ? total = 3024 + 672 + 6 + 32
 = 3734
 
Q8: In an examination of Mathematics paper, there are 20 questions of equal marks and the question 
paper is divided into three sections : ?? , ?? and ?? . A student is required to attempt total 15 questions 
taking at least 4 questions from each section. If section ?? has 8 questions, section ?? has 6 questions 
and section ?? has 6 questions, then the total number of ways a student can select 15 questions is 
[JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: 11376 
If 4 questions from each section are selected 
Remaining 3 questions can be selected either in (1,1,1) or (3,0,0) or (2,1,0) 
 ? Total ways = 
8
c
5
· 
6
c
5
· 
6
c
5
+ 
8
c
6
 
6
c
5
· 
6
c
4
× 2 +
 
8
c
5
· 
6
c
6
· 
6
c
4
× 2 + 
8
c
4
· 
6
c
6
· 
6
c
5
× 2 + 
8
c
7
· 
6
c
4
· 
6
c
4
 = 56.6.6 + 28.6.15.2 + 56.15.2 + 70.6.2
 +8.15.15
 = 2016 + 5040 + 1680 + 840 + 1800 = 11376
 
Q9: If 
 
????
?? ?? ?? +
 
????
?? ?? ?? + ? +
 
????
?? ?? ????
=
?? ?? with ?????? (?? , ?? ) = ?? , then ?? + ?? is equal to    [JEE Main 2024 
(Online) 29th January Morning Shift] 
Ans: 2041 
Explanation: Currently no explanation available 
 
 
 
 
2023 
Q1: The total number of three-digit numbers, divisible by 3, which can be formed using the digits 1 , 3 
, 5 , 8 , if repetition of digits is allowed, is: 
(a) 21 
(b) 22 
(c) 18 
(d) 20 
Ans: (b) 
The number of three-digit numbers divisible by 3 by considering the possible sums of digits that are 
divisible by 3. Your approach is as follows: 
1. Sum of digits is 3: (1 , 1 , 1) - 1 possible number 
2. Sum of digits is 9: (1 , 3 , 5) and (3 , 3 , 3) - 
Let's consider the cases separately: 
a. Sum of digits is 9: (1 , 3 , 5) For this case, we can arrange the digits in 3! ways : 135, 153, 315, 351, 
513, and 531. 
b. Sum of digits is 9: (3 , 3 , 3) For this case, since all the digits are the same, there is only 1 possible 
number : 333. 
Now, the total number of possible numbers when the sum of digits is 9 is : 
3 ! + 1 = 6 + 1 = 7 
3. Sum of digits is 12: (1 , 3 , 8) - 3! possible numbers 
4. Sum of digits is 15: (5 , 5 , 5) - 1 possible number 
5. Sum of digits is 18: (5 , 5 , 8) - 3!/2! possible numbers (since 5 is repeated) 
6. Sum of digits is 21: (5 , 8 , 8) - 3!/2! possible numbers (since 8 is repeated) 
7. Sum of digits is 24: (8 , 8 , 8) - 1 possible number 
Adding up the possible numbers for each case, we get: 
 
 So, there are a total of 22 three-digit numbers divisible by 3 that can be formed using the digits 1, 3, 5, 8 
with repetition allowed. 
 
Q2: All words, with or without meaning, are made using all the letters of the word MONDAY. These 
words are written as in a dictionary with serial numbers. The serial number of the word MONDAY is: 
(a) 324 
(b) 328 
(c) 326 
(d) 327 
Ans: (d) 
Apologies for the confusion earlier. Let's correct the calculation: 
The letters of "MONDAY" arranged in alphabetical order are: A, D, M, N, O, Y. 
Fix A as the first letter, we can arrange the remaining 5 letters in 5! ways = 120 ways. 
Fix D as the first letter, we can arrange the remaining 5 letters in 5! ways = 120 ways. 
So far, we have 120 (for A) + 120 (for D) = 240 words. 
Now, let's proceed with words starting with M: 
Fix MA as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways. 
Fix MD as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways. 
Fix MN as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways.