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JEE Main Previous Year Questions (2025): Permutations and Combinations
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Page 1
JEE Main Previous Year Questions
(2025): Permutations and
Combinations
Q1: The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys
sit together or no two boys sit together, is ____ .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 17280
Solution:
A : number of ways that all boys sit together = 5! × 5!
B : number of ways if no 2 boys
sit together = 4! × 5!
A n B = ??
Required no. of ways = 5! × 5! + 4! × 5! = 17280
Q2: The number of 3 -digit numbers, that are divisible by 2 and 3 , but not divisible by
4 and 9 , is ____
JEE Main 2025 (Online) 24th January Morning Shift
Solution:
No, of 3 digits = 999 - 99 = 900
No. of 3 digit numbers divisible by 2&3 i.e. by 6
900
6
= 150
No. of 3 digit numbers divisible by 4&9 i.e. by 36
900
36
= 25
? No of 3 digit numbers divisible by 2&3 but not by 4&9
150 - 25 = 125
Q3: Number of functions ?? : {?? , ?? , … , ?????? } ? {?? , ?? }, that assign 1 to exactly one of the
positive integers less than or equal to 98 , is equal to ____ .
JEE Main 2025 (Online) 24th January Evening Shift
Ans: 392
Solution:
Page 2
JEE Main Previous Year Questions
(2025): Permutations and
Combinations
Q1: The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys
sit together or no two boys sit together, is ____ .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 17280
Solution:
A : number of ways that all boys sit together = 5! × 5!
B : number of ways if no 2 boys
sit together = 4! × 5!
A n B = ??
Required no. of ways = 5! × 5! + 4! × 5! = 17280
Q2: The number of 3 -digit numbers, that are divisible by 2 and 3 , but not divisible by
4 and 9 , is ____
JEE Main 2025 (Online) 24th January Morning Shift
Solution:
No, of 3 digits = 999 - 99 = 900
No. of 3 digit numbers divisible by 2&3 i.e. by 6
900
6
= 150
No. of 3 digit numbers divisible by 4&9 i.e. by 36
900
36
= 25
? No of 3 digit numbers divisible by 2&3 but not by 4&9
150 - 25 = 125
Q3: Number of functions ?? : {?? , ?? , … , ?????? } ? {?? , ?? }, that assign 1 to exactly one of the
positive integers less than or equal to 98 , is equal to ____ .
JEE Main 2025 (Online) 24th January Evening Shift
Ans: 392
Solution:
392 Ans.
Q4: The number of natural numbers, between 212 and 999, such that the sum of their
digits is 15 , is ____ .
JEE Main 2025 (Online) 28th January Evening Shift
Ans: 64
Solution:
Let x = 2 ? y + z = 13
(4,9), (5,8), (6,7), (7,6), (8,5), (9,4), ? 6
Let ?? = 3 ? ?? + ?? = 12
(3,9), (4,8), … … . , (9,3) ? 7
Let ?? = 4 ? ?? + ?? = 11
(2,9), (3,8), … … … , (9,1) ? 9
Let ?? = 5 ? ?? + ?? = 10
(1,9), (2,8), … … … , (9,1) ? 10
Let ?? = 6 ? ?? + ?? = 9
(0,9), (1,8), … … . , (9,0) ? 9
Let x = 7 ? y + z = 8
(0,9), (1,7), … … … , (8,0) ? 9
Let ?? = 8 ? ?? + ?? = 7
(0,7), (1,6), … … … , (7,0) ? 8
Let ?? = 9 ? ?? + ?? = 6
(0,6), (1,5), … … … , (6,0) ? 7
Total = 6 = 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64
Q5: The number of 6-letter words, with or without meaning, that can be formed using
the letters of the word MATHS such that any letter that appears in the word must
appear at least twice, is ____ .
JEE Main 2025 (Online) 29th January Morning Shift
Ans: 1405
Solution:
(i) Single letter is used, then no. of words = 5
(ii) Two distinct letters are used, then no. of words
5
C
2
× (
6!
2! 4!
× 2 +
6!
3! 3!
) = 10(30 + 20) = 500
(iii) Three distinct letters are used, then no. of words
Page 3
JEE Main Previous Year Questions
(2025): Permutations and
Combinations
Q1: The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys
sit together or no two boys sit together, is ____ .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 17280
Solution:
A : number of ways that all boys sit together = 5! × 5!
B : number of ways if no 2 boys
sit together = 4! × 5!
A n B = ??
Required no. of ways = 5! × 5! + 4! × 5! = 17280
Q2: The number of 3 -digit numbers, that are divisible by 2 and 3 , but not divisible by
4 and 9 , is ____
JEE Main 2025 (Online) 24th January Morning Shift
Solution:
No, of 3 digits = 999 - 99 = 900
No. of 3 digit numbers divisible by 2&3 i.e. by 6
900
6
= 150
No. of 3 digit numbers divisible by 4&9 i.e. by 36
900
36
= 25
? No of 3 digit numbers divisible by 2&3 but not by 4&9
150 - 25 = 125
Q3: Number of functions ?? : {?? , ?? , … , ?????? } ? {?? , ?? }, that assign 1 to exactly one of the
positive integers less than or equal to 98 , is equal to ____ .
JEE Main 2025 (Online) 24th January Evening Shift
Ans: 392
Solution:
392 Ans.
Q4: The number of natural numbers, between 212 and 999, such that the sum of their
digits is 15 , is ____ .
JEE Main 2025 (Online) 28th January Evening Shift
Ans: 64
Solution:
Let x = 2 ? y + z = 13
(4,9), (5,8), (6,7), (7,6), (8,5), (9,4), ? 6
Let ?? = 3 ? ?? + ?? = 12
(3,9), (4,8), … … . , (9,3) ? 7
Let ?? = 4 ? ?? + ?? = 11
(2,9), (3,8), … … … , (9,1) ? 9
Let ?? = 5 ? ?? + ?? = 10
(1,9), (2,8), … … … , (9,1) ? 10
Let ?? = 6 ? ?? + ?? = 9
(0,9), (1,8), … … . , (9,0) ? 9
Let x = 7 ? y + z = 8
(0,9), (1,7), … … … , (8,0) ? 9
Let ?? = 8 ? ?? + ?? = 7
(0,7), (1,6), … … … , (7,0) ? 8
Let ?? = 9 ? ?? + ?? = 6
(0,6), (1,5), … … … , (6,0) ? 7
Total = 6 = 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64
Q5: The number of 6-letter words, with or without meaning, that can be formed using
the letters of the word MATHS such that any letter that appears in the word must
appear at least twice, is ____ .
JEE Main 2025 (Online) 29th January Morning Shift
Ans: 1405
Solution:
(i) Single letter is used, then no. of words = 5
(ii) Two distinct letters are used, then no. of words
5
C
2
× (
6!
2! 4!
× 2 +
6!
3! 3!
) = 10(30 + 20) = 500
(iii) Three distinct letters are used, then no. of words
5
C
3
×
6!
2! 2! 2!
= 900
Total no. of words = 1405
Q6: If the number of seven-digit numbers, such that the sum of their digits is even, is
?? · ?? · ????
?? ; ?? , ?? ? {?? , ?? , ?? , … , ?? }, then ?? + ?? is equal to ____
JEE Main 2025 (Online) 3rd April Morning Shift
Ans: 14
Solution:
When numbers are uniformly distributed, half of them have even digit sums and half have odd
digits sums.
Number of 7-digit numbers with even digit sum =
1
2
· 9 · 10
6
= 4.5 · 10
6
Q7: All five letter words are made using all the letters ?? , ?? , ?? , ?? , ?? and arranged as in
an English dictionary with serial numbers. Let the word at serial number ?? be denoted
by ?? ?? . Let the probability ?? (?? ?? ) of choosing the word ?? ?? satisfy ?? (?? ?? ) =
???? (?? ?? -?? ), ?? > ?? .
If ?? (?????????? ) =
?? ?? ?? ?? -?? , ?? , ?? ? N, then ?? + ?? is equal to : ____
JEE Main 2025 (Online) 3rd April Morning Shift
Ans: 183
Solution:
Firstly, by this rule, we note:
?? (?? 1
) = ??
?? (?? 2
) = 2??
?? (?? 3
) = 4??
?? (?? ?? ) = 2
?? -1
??
To find the initial probability ?? , consider the total probability must sum to 1 across all 120
possible words (since 5! = 120 ):
?
?? =1
120
??? (?? ?? ) = 1
This is a geometric series sum where:
?? (1 + 2 + 2
2
+ ? + 2
119
) = 1
Since the series sum 1 + 2 + 2
2
+ ? + 2
119
is equal to 2
120
- 1, we have:
?? (2
120
- 1) = 1 ? ?? =
1
2
120
- 1
Thus, the probability for the ?? -th word is:
?? (?? ?? ) =
2
?? -1
2
120
- 1
(?? )
Page 4
JEE Main Previous Year Questions
(2025): Permutations and
Combinations
Q1: The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys
sit together or no two boys sit together, is ____ .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 17280
Solution:
A : number of ways that all boys sit together = 5! × 5!
B : number of ways if no 2 boys
sit together = 4! × 5!
A n B = ??
Required no. of ways = 5! × 5! + 4! × 5! = 17280
Q2: The number of 3 -digit numbers, that are divisible by 2 and 3 , but not divisible by
4 and 9 , is ____
JEE Main 2025 (Online) 24th January Morning Shift
Solution:
No, of 3 digits = 999 - 99 = 900
No. of 3 digit numbers divisible by 2&3 i.e. by 6
900
6
= 150
No. of 3 digit numbers divisible by 4&9 i.e. by 36
900
36
= 25
? No of 3 digit numbers divisible by 2&3 but not by 4&9
150 - 25 = 125
Q3: Number of functions ?? : {?? , ?? , … , ?????? } ? {?? , ?? }, that assign 1 to exactly one of the
positive integers less than or equal to 98 , is equal to ____ .
JEE Main 2025 (Online) 24th January Evening Shift
Ans: 392
Solution:
392 Ans.
Q4: The number of natural numbers, between 212 and 999, such that the sum of their
digits is 15 , is ____ .
JEE Main 2025 (Online) 28th January Evening Shift
Ans: 64
Solution:
Let x = 2 ? y + z = 13
(4,9), (5,8), (6,7), (7,6), (8,5), (9,4), ? 6
Let ?? = 3 ? ?? + ?? = 12
(3,9), (4,8), … … . , (9,3) ? 7
Let ?? = 4 ? ?? + ?? = 11
(2,9), (3,8), … … … , (9,1) ? 9
Let ?? = 5 ? ?? + ?? = 10
(1,9), (2,8), … … … , (9,1) ? 10
Let ?? = 6 ? ?? + ?? = 9
(0,9), (1,8), … … . , (9,0) ? 9
Let x = 7 ? y + z = 8
(0,9), (1,7), … … … , (8,0) ? 9
Let ?? = 8 ? ?? + ?? = 7
(0,7), (1,6), … … … , (7,0) ? 8
Let ?? = 9 ? ?? + ?? = 6
(0,6), (1,5), … … … , (6,0) ? 7
Total = 6 = 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64
Q5: The number of 6-letter words, with or without meaning, that can be formed using
the letters of the word MATHS such that any letter that appears in the word must
appear at least twice, is ____ .
JEE Main 2025 (Online) 29th January Morning Shift
Ans: 1405
Solution:
(i) Single letter is used, then no. of words = 5
(ii) Two distinct letters are used, then no. of words
5
C
2
× (
6!
2! 4!
× 2 +
6!
3! 3!
) = 10(30 + 20) = 500
(iii) Three distinct letters are used, then no. of words
5
C
3
×
6!
2! 2! 2!
= 900
Total no. of words = 1405
Q6: If the number of seven-digit numbers, such that the sum of their digits is even, is
?? · ?? · ????
?? ; ?? , ?? ? {?? , ?? , ?? , … , ?? }, then ?? + ?? is equal to ____
JEE Main 2025 (Online) 3rd April Morning Shift
Ans: 14
Solution:
When numbers are uniformly distributed, half of them have even digit sums and half have odd
digits sums.
Number of 7-digit numbers with even digit sum =
1
2
· 9 · 10
6
= 4.5 · 10
6
Q7: All five letter words are made using all the letters ?? , ?? , ?? , ?? , ?? and arranged as in
an English dictionary with serial numbers. Let the word at serial number ?? be denoted
by ?? ?? . Let the probability ?? (?? ?? ) of choosing the word ?? ?? satisfy ?? (?? ?? ) =
???? (?? ?? -?? ), ?? > ?? .
If ?? (?????????? ) =
?? ?? ?? ?? -?? , ?? , ?? ? N, then ?? + ?? is equal to : ____
JEE Main 2025 (Online) 3rd April Morning Shift
Ans: 183
Solution:
Firstly, by this rule, we note:
?? (?? 1
) = ??
?? (?? 2
) = 2??
?? (?? 3
) = 4??
?? (?? ?? ) = 2
?? -1
??
To find the initial probability ?? , consider the total probability must sum to 1 across all 120
possible words (since 5! = 120 ):
?
?? =1
120
??? (?? ?? ) = 1
This is a geometric series sum where:
?? (1 + 2 + 2
2
+ ? + 2
119
) = 1
Since the series sum 1 + 2 + 2
2
+ ? + 2
119
is equal to 2
120
- 1, we have:
?? (2
120
- 1) = 1 ? ?? =
1
2
120
- 1
Thus, the probability for the ?? -th word is:
?? (?? ?? ) =
2
?? -1
2
120
- 1
(?? )
Next, determine the position of "CDBEA". Starting from the first letter:
Words beginning with ' ?? ' : 4! = 24
Words beginning with ' ?? ': 4! = 24
Words beginning with ' ?? ':
????
*
: 3! = 6
????
*
: 3! = 6
CDA** : 2! = 2
CDBA
*
: 1! = 1
Summing these, the position of "CDBEA" is the 64th word.
Substitute into equation (i):
?? (CDBEA) = ?? (?? 64
) =
2
63
2
120
- 1
Given ?? (CDBEA) =
2
?? 2
?? -1
, we find:
?? = 63
?? = 120
Thus, the sum ?? + ?? = 63 + 120 = 183.
Q8: Let ?? and ?? , (?? < ?? ), be two 2 -digit numbers. Then the total numbers of pairs
(?? , ?? ), such that ?????? (?? , ?? ) = ?? , is ____ .
JEE Main 2025 (Online) 4th April Evening Shift
Ans: 64
Solution:
?? = 6?? , ?? = 6??
So gcd(?? , ?? ) = 6 ? gcd(?? , ?? ) = 1
?? = 6?? = 10 ? ?? = [
10
6
] = 2
?? = 6?? = 99 ? ?? = [
99
6
] = 16
So ?? , ?? ? {2,3, … ,16}, and we count how many coprime pairs ( ?? , ?? ) with ?? < ?? , gcd(?? , ?? ) = 1
?? = 2 ? ?? = 3,5,7,9,11,13,15 ? 7
?? = 3 ? b = 4,5,7,8,10,11,13,14,16 ? 9
?? = 4 ? ?? = 5,7,9,11,13,15 ? 6
?? = 5 ? b = 6,7,8,9,11,12,13,14,16 ? 9
?? = 6 ? b = 7,11,13 ? 3
?? = 7 ? ?? = 8,9,10,11,12,13,15,16 ? 8
?? = 8 ? b = 9,11,13,15 ? 4
?? = 9 ? b = 10,11,13,14,16 ? 5
?? = 10 ? ?? = 11,13 ? 2
?? = 11 ? ?? = 12,13,14,15,16 ? 5
?? = 12 ? ?? = 13,17 ×? only 13 is valid ? 1
?? = 13 ? ?? = 14,15,16 ? 3
?? = 14 ? ?? = 15, ? 1
?? = 15 ? ?? = 16 ? 1
Page 5
JEE Main Previous Year Questions
(2025): Permutations and
Combinations
Q1: The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys
sit together or no two boys sit together, is ____ .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 17280
Solution:
A : number of ways that all boys sit together = 5! × 5!
B : number of ways if no 2 boys
sit together = 4! × 5!
A n B = ??
Required no. of ways = 5! × 5! + 4! × 5! = 17280
Q2: The number of 3 -digit numbers, that are divisible by 2 and 3 , but not divisible by
4 and 9 , is ____
JEE Main 2025 (Online) 24th January Morning Shift
Solution:
No, of 3 digits = 999 - 99 = 900
No. of 3 digit numbers divisible by 2&3 i.e. by 6
900
6
= 150
No. of 3 digit numbers divisible by 4&9 i.e. by 36
900
36
= 25
? No of 3 digit numbers divisible by 2&3 but not by 4&9
150 - 25 = 125
Q3: Number of functions ?? : {?? , ?? , … , ?????? } ? {?? , ?? }, that assign 1 to exactly one of the
positive integers less than or equal to 98 , is equal to ____ .
JEE Main 2025 (Online) 24th January Evening Shift
Ans: 392
Solution:
392 Ans.
Q4: The number of natural numbers, between 212 and 999, such that the sum of their
digits is 15 , is ____ .
JEE Main 2025 (Online) 28th January Evening Shift
Ans: 64
Solution:
Let x = 2 ? y + z = 13
(4,9), (5,8), (6,7), (7,6), (8,5), (9,4), ? 6
Let ?? = 3 ? ?? + ?? = 12
(3,9), (4,8), … … . , (9,3) ? 7
Let ?? = 4 ? ?? + ?? = 11
(2,9), (3,8), … … … , (9,1) ? 9
Let ?? = 5 ? ?? + ?? = 10
(1,9), (2,8), … … … , (9,1) ? 10
Let ?? = 6 ? ?? + ?? = 9
(0,9), (1,8), … … . , (9,0) ? 9
Let x = 7 ? y + z = 8
(0,9), (1,7), … … … , (8,0) ? 9
Let ?? = 8 ? ?? + ?? = 7
(0,7), (1,6), … … … , (7,0) ? 8
Let ?? = 9 ? ?? + ?? = 6
(0,6), (1,5), … … … , (6,0) ? 7
Total = 6 = 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64
Q5: The number of 6-letter words, with or without meaning, that can be formed using
the letters of the word MATHS such that any letter that appears in the word must
appear at least twice, is ____ .
JEE Main 2025 (Online) 29th January Morning Shift
Ans: 1405
Solution:
(i) Single letter is used, then no. of words = 5
(ii) Two distinct letters are used, then no. of words
5
C
2
× (
6!
2! 4!
× 2 +
6!
3! 3!
) = 10(30 + 20) = 500
(iii) Three distinct letters are used, then no. of words
5
C
3
×
6!
2! 2! 2!
= 900
Total no. of words = 1405
Q6: If the number of seven-digit numbers, such that the sum of their digits is even, is
?? · ?? · ????
?? ; ?? , ?? ? {?? , ?? , ?? , … , ?? }, then ?? + ?? is equal to ____
JEE Main 2025 (Online) 3rd April Morning Shift
Ans: 14
Solution:
When numbers are uniformly distributed, half of them have even digit sums and half have odd
digits sums.
Number of 7-digit numbers with even digit sum =
1
2
· 9 · 10
6
= 4.5 · 10
6
Q7: All five letter words are made using all the letters ?? , ?? , ?? , ?? , ?? and arranged as in
an English dictionary with serial numbers. Let the word at serial number ?? be denoted
by ?? ?? . Let the probability ?? (?? ?? ) of choosing the word ?? ?? satisfy ?? (?? ?? ) =
???? (?? ?? -?? ), ?? > ?? .
If ?? (?????????? ) =
?? ?? ?? ?? -?? , ?? , ?? ? N, then ?? + ?? is equal to : ____
JEE Main 2025 (Online) 3rd April Morning Shift
Ans: 183
Solution:
Firstly, by this rule, we note:
?? (?? 1
) = ??
?? (?? 2
) = 2??
?? (?? 3
) = 4??
?? (?? ?? ) = 2
?? -1
??
To find the initial probability ?? , consider the total probability must sum to 1 across all 120
possible words (since 5! = 120 ):
?
?? =1
120
??? (?? ?? ) = 1
This is a geometric series sum where:
?? (1 + 2 + 2
2
+ ? + 2
119
) = 1
Since the series sum 1 + 2 + 2
2
+ ? + 2
119
is equal to 2
120
- 1, we have:
?? (2
120
- 1) = 1 ? ?? =
1
2
120
- 1
Thus, the probability for the ?? -th word is:
?? (?? ?? ) =
2
?? -1
2
120
- 1
(?? )
Next, determine the position of "CDBEA". Starting from the first letter:
Words beginning with ' ?? ' : 4! = 24
Words beginning with ' ?? ': 4! = 24
Words beginning with ' ?? ':
????
*
: 3! = 6
????
*
: 3! = 6
CDA** : 2! = 2
CDBA
*
: 1! = 1
Summing these, the position of "CDBEA" is the 64th word.
Substitute into equation (i):
?? (CDBEA) = ?? (?? 64
) =
2
63
2
120
- 1
Given ?? (CDBEA) =
2
?? 2
?? -1
, we find:
?? = 63
?? = 120
Thus, the sum ?? + ?? = 63 + 120 = 183.
Q8: Let ?? and ?? , (?? < ?? ), be two 2 -digit numbers. Then the total numbers of pairs
(?? , ?? ), such that ?????? (?? , ?? ) = ?? , is ____ .
JEE Main 2025 (Online) 4th April Evening Shift
Ans: 64
Solution:
?? = 6?? , ?? = 6??
So gcd(?? , ?? ) = 6 ? gcd(?? , ?? ) = 1
?? = 6?? = 10 ? ?? = [
10
6
] = 2
?? = 6?? = 99 ? ?? = [
99
6
] = 16
So ?? , ?? ? {2,3, … ,16}, and we count how many coprime pairs ( ?? , ?? ) with ?? < ?? , gcd(?? , ?? ) = 1
?? = 2 ? ?? = 3,5,7,9,11,13,15 ? 7
?? = 3 ? b = 4,5,7,8,10,11,13,14,16 ? 9
?? = 4 ? ?? = 5,7,9,11,13,15 ? 6
?? = 5 ? b = 6,7,8,9,11,12,13,14,16 ? 9
?? = 6 ? b = 7,11,13 ? 3
?? = 7 ? ?? = 8,9,10,11,12,13,15,16 ? 8
?? = 8 ? b = 9,11,13,15 ? 4
?? = 9 ? b = 10,11,13,14,16 ? 5
?? = 10 ? ?? = 11,13 ? 2
?? = 11 ? ?? = 12,13,14,15,16 ? 5
?? = 12 ? ?? = 13,17 ×? only 13 is valid ? 1
?? = 13 ? ?? = 14,15,16 ? 3
?? = 14 ? ?? = 15, ? 1
?? = 15 ? ?? = 16 ? 1
Total = 7 + 9 + 6 + 9 + 3 + 8 + 4 + 5 + 2 + 5 + 1
+3 + 1 + 1 = 64
Q9: From all the English alphabets, five letters are chosen and are arranged in
alphabetical order. The total number of ways, in which the middle letter is ' ?? ', is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 6084
B. 5148
C. 14950
D. 4356
Ans: B
Solution:
First, note that we are choosing ?? distinct letters (in strictly increasing alphabetical order) such
that the middle (third) letter is ' ?? '. Symbolically, if we denote the chosen letters as:
?? 1
< ?? 2
< ?? 3
< ?? 4
< ?? 5
,
we want ?? 3
= M. The English alphabet has 26 letters, and M is the 13
th
.
Step 1: Letters before M
The letters before M are {?? , ?? , ?? , … , ?? }.
There are 12 letters here ( ?? through ?? ).
We need to pick 2 of these 12 letters to occupy ?? 1
and ?? 2
.
The number of ways to choose 2 letters out of 12 is
12
C
2
.
Step 2: Letters after M
The letters after M are {?? , ?? , ?? , … , ?? }.
There are 13 letters here ( ?? through ?? ).
We need to pick 2 of these 13 letters to occupy ?? 4
and ?? 5
.
The number of ways to choose 2 letters out of 13 is
13
C
2
.
Step 3: Multiply the choices
Since these choices are independent (picking the two letters before M and two letters after M ),
the total number of ways is:
12
C
2
×
13
C
2
Calculate each combination:
12
C
2
=
12×11
2
= 66,
13
C
2
=
13×12
2
= 78.
So,
12
C
2
×
13
C
2
= 66 × 78 = 5148.
Read MoreFAQs on JEE Main Previous Year Questions (2025): Permutations and Combinations
| 1. What are permutations and combinations, and how do they differ? |  |
Ans. Permutations and combinations are both methods used in counting and arranging objects. Permutations refer to the different arrangements of a set of items where the order matters. For example, the arrangements of the letters A, B, and C (ABC, ACB, BAC, BCA, CAB, CBA) are permutations. On the other hand, combinations are selections of items where the order does not matter. For instance, choosing 2 letters from A, B, and C (AB, AC, BC) are combinations. Thus, while permutations focus on the arrangement, combinations focus on the selection.
| 2. How is the formula for permutations calculated? | |
Ans. The formula for calculating permutations of n items taken r at a time is given by nPr = n! / (n - r)!. Here, n! (n factorial) is the product of all positive integers up to n. This formula counts the number of ways to arrange r items selected from n total items, considering the order of selection.
| 3. What is the formula for combinations, and when is it used? | |
Ans. The formula for combinations of n items taken r at a time is given by nCr = n! / [r! * (n - r)!]. This formula is used when the order of selection does not matter, and it calculates the number of ways to choose r items from a set of n items without regard to the sequence in which they are selected.
| 4. Can you provide an example of a real-life application of permutations and combinations? | |
Ans. One real-life application of permutations and combinations is in organizing a tournament. For example, if a tournament has 8 teams and we want to know the different ways to arrange matches between them, we can use permutations. Conversely, if we want to select 4 teams to form a committee from those 8, we would use combinations. This helps in planning and structuring competitive events efficiently.
| 5. How can I improve my skills in solving problems related to permutations and combinations? | |
Ans. To improve your skills in solving problems related to permutations and combinations, practice is key. Start by understanding the basic concepts and formulas. Work through a variety of problems, starting from simple to more complex ones. Utilize resources such as textbooks, online tutorials, and previous exam papers. Additionally, joining study groups or forums can provide support and different perspectives on tackling problems.
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