Permutations and Combinations: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Permutations and Combinations  
 
(January 2026) 
 
 
Q1: Three persons enter in a lift at the ground floor. The lift will go up to 10th floor. The 
number of ways, in which the three persons can exit the lift at three different floors, if the 
lift does not stop at first, second and third floors, is equal to ________. 
Ans: 210 
Sol: 
The lift goes up to the 10
th
 floor. 
The lift does not stop at 1
st
, 2
nd
, 3
rd
 floors. 
So, the persons can get down only at the remaining floors. 
4, 5, 6, 7, 8, 9, 10 ? 7 floors 
Now, all three persons must exit at three different floors (no two can get down at the same floor). 
Also, the three persons are distinct, so who gets down where matters. 
First, choose which 3 floors (out of 7) will be used: 
 
Next, assign (arrange) the 3 distinct persons to these 3 chosen floors: 
Number of arrangements of 3 persons = 3! = 6 
So, total number of ways: 
Total ways = 7C 3 × 3! = 35 × 6 = 210 
 
Q2: The number of numbers greater than 5000, less than 9000 and divisible by 3, that can 
be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is 
Ans: 42 
Sol: 
To find numbers of number greater than 5000 and less than 9000 so number will start with 5 or 
9 of the form 5abc 
and a, b, c ? {0, 1, 2, 5, 9} also the number is divisible by 3. 
This means 5 + a + b + c is divisible by 3 
so the sum a + b + c is of the type 3n + 1. 
unit place c determined by the sum of previous 3 places. 
rem(a + b) = remainder of a + b 
rem(a+b) combination (a,b) value of c total = combination (a, b) × c(choice) 
0 (0, 0), (0, 9), (9, 0), (9, 9), (1, 2), (2, 1), (5, 
1), (1, 5) 
1 8 × 1 = 8 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Permutations and Combinations  
 
(January 2026) 
 
 
Q1: Three persons enter in a lift at the ground floor. The lift will go up to 10th floor. The 
number of ways, in which the three persons can exit the lift at three different floors, if the 
lift does not stop at first, second and third floors, is equal to ________. 
Ans: 210 
Sol: 
The lift goes up to the 10
th
 floor. 
The lift does not stop at 1
st
, 2
nd
, 3
rd
 floors. 
So, the persons can get down only at the remaining floors. 
4, 5, 6, 7, 8, 9, 10 ? 7 floors 
Now, all three persons must exit at three different floors (no two can get down at the same floor). 
Also, the three persons are distinct, so who gets down where matters. 
First, choose which 3 floors (out of 7) will be used: 
 
Next, assign (arrange) the 3 distinct persons to these 3 chosen floors: 
Number of arrangements of 3 persons = 3! = 6 
So, total number of ways: 
Total ways = 7C 3 × 3! = 35 × 6 = 210 
 
Q2: The number of numbers greater than 5000, less than 9000 and divisible by 3, that can 
be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is 
Ans: 42 
Sol: 
To find numbers of number greater than 5000 and less than 9000 so number will start with 5 or 
9 of the form 5abc 
and a, b, c ? {0, 1, 2, 5, 9} also the number is divisible by 3. 
This means 5 + a + b + c is divisible by 3 
so the sum a + b + c is of the type 3n + 1. 
unit place c determined by the sum of previous 3 places. 
rem(a + b) = remainder of a + b 
rem(a+b) combination (a,b) value of c total = combination (a, b) × c(choice) 
0 (0, 0), (0, 9), (9, 0), (9, 9), (1, 2), (2, 1), (5, 
1), (1, 5) 
1 8 × 1 = 8 
1 (0, 1), (9, 1), (1, 0), (1, 9), (2, 2), (2, 5), (5, 
2), (5, 5) 
0, 9 8 × 2 = 16 
2 (0, 2), (0, 5), (9, 2), (9, 5), (2, 0), (2, 9), (5, 
0), (5, 9), (1, 1) 
2, 5 9 × 2 = 18 
Total = 8 + 16 + 18 = 42 
 
 
Q3: Let S denote the set of 4-digit numbers abcd such that a > b > c > d and P denote the 
set of 5 -digit numbers having product of its digits equal to 20. Then n(S) + n(P) is equal 
to 
Ans: 260 
Sol: 
For set S: 4-digit numbers abcd with a > b > c > d 
digits are chosen from {0, 1, 2, …, 9} 
Since order is strictly fixed, any 4 distinct digits chosen will form exactly one such number. 
Number of ways to choose 4 digits from 10 is ¹ °C 4 
 
For set P: 5-digit numbers with product of digits = 20 
Possible sets of digits (using digits 1-9): 
Case 1: {5, 4, 1, 1, 1} 
Number of arrangements  
Case 2: {5, 2, 2, 1, 1} 
Number of arrangements  
n(P) = 20 + 30 = 50 
Calculating final value: 
n(S) + n(P) = 210 + 50 = 260 
The value of n(S) + n(P) is 260. 
 
 
Q4: The number of 4 -letter words, with or without meaning, which can be formed using 
the letters PQRPQRSTUVP, is _____. 
Ans: 1422 
Sol: 
Letter frequency 
P : 3 
R, Q : 2 
S, T, U, V : 1 
4 letter words can be of type 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Permutations and Combinations  
 
(January 2026) 
 
 
Q1: Three persons enter in a lift at the ground floor. The lift will go up to 10th floor. The 
number of ways, in which the three persons can exit the lift at three different floors, if the 
lift does not stop at first, second and third floors, is equal to ________. 
Ans: 210 
Sol: 
The lift goes up to the 10
th
 floor. 
The lift does not stop at 1
st
, 2
nd
, 3
rd
 floors. 
So, the persons can get down only at the remaining floors. 
4, 5, 6, 7, 8, 9, 10 ? 7 floors 
Now, all three persons must exit at three different floors (no two can get down at the same floor). 
Also, the three persons are distinct, so who gets down where matters. 
First, choose which 3 floors (out of 7) will be used: 
 
Next, assign (arrange) the 3 distinct persons to these 3 chosen floors: 
Number of arrangements of 3 persons = 3! = 6 
So, total number of ways: 
Total ways = 7C 3 × 3! = 35 × 6 = 210 
 
Q2: The number of numbers greater than 5000, less than 9000 and divisible by 3, that can 
be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is 
Ans: 42 
Sol: 
To find numbers of number greater than 5000 and less than 9000 so number will start with 5 or 
9 of the form 5abc 
and a, b, c ? {0, 1, 2, 5, 9} also the number is divisible by 3. 
This means 5 + a + b + c is divisible by 3 
so the sum a + b + c is of the type 3n + 1. 
unit place c determined by the sum of previous 3 places. 
rem(a + b) = remainder of a + b 
rem(a+b) combination (a,b) value of c total = combination (a, b) × c(choice) 
0 (0, 0), (0, 9), (9, 0), (9, 9), (1, 2), (2, 1), (5, 
1), (1, 5) 
1 8 × 1 = 8 
1 (0, 1), (9, 1), (1, 0), (1, 9), (2, 2), (2, 5), (5, 
2), (5, 5) 
0, 9 8 × 2 = 16 
2 (0, 2), (0, 5), (9, 2), (9, 5), (2, 0), (2, 9), (5, 
0), (5, 9), (1, 1) 
2, 5 9 × 2 = 18 
Total = 8 + 16 + 18 = 42 
 
 
Q3: Let S denote the set of 4-digit numbers abcd such that a > b > c > d and P denote the 
set of 5 -digit numbers having product of its digits equal to 20. Then n(S) + n(P) is equal 
to 
Ans: 260 
Sol: 
For set S: 4-digit numbers abcd with a > b > c > d 
digits are chosen from {0, 1, 2, …, 9} 
Since order is strictly fixed, any 4 distinct digits chosen will form exactly one such number. 
Number of ways to choose 4 digits from 10 is ¹ °C 4 
 
For set P: 5-digit numbers with product of digits = 20 
Possible sets of digits (using digits 1-9): 
Case 1: {5, 4, 1, 1, 1} 
Number of arrangements  
Case 2: {5, 2, 2, 1, 1} 
Number of arrangements  
n(P) = 20 + 30 = 50 
Calculating final value: 
n(S) + n(P) = 210 + 50 = 260 
The value of n(S) + n(P) is 260. 
 
 
Q4: The number of 4 -letter words, with or without meaning, which can be formed using 
the letters PQRPQRSTUVP, is _____. 
Ans: 1422 
Sol: 
Letter frequency 
P : 3 
R, Q : 2 
S, T, U, V : 1 
4 letter words can be of type 
? ABCD or AABC, AABB, AAAB 
 
= 1422 
 
 
Q5: Let ABC be a triangle. Consider four points p
1
, p
2
, p
3
, p
4
 on the side AB, five points p
5
, 
p
6
, p
7
, p
8
, p
9
 on the side BC, and four points p
10
, p
11
, p
12
, p
13
 on the side AC. None of these 
points is a vertex of the triangle ABC. Then the total number of pentagons, that can be 
formed by taking all the vertices from the points p
1
, p
2
 , …, p
13
, is 
Ans: 660 
Sol: 
4C 2 · 5C 2 · 4C 1 + 4C 2 · 5C 2 · 4C 1 + 4C 2 · 4C 2 · 5C 1 = 240 + 240 + 180 = 660 
 
Q6: Let S = {(m, n) : m, n ? {1, 2, 3, …, 50}}. If the number of elements (m, n) in S such 
that 6
m
 + 9
n
 is a multiple of 5 is p and the number of elements (m, n) in S such that m + n 
is a square of a prime number is q, then p + q is equal to _____. 
Ans: 1333 
Sol: 
5 | 6
m
 + 9
n 
? 5 | 1
m
 + (-1)
n 
? m and n has to be opposite parity. 
²C 1 × ² 5C 1 · ² 5C 1 = 625 × 2 = 1250 
For m + n = K² for some prime K. 
m + n ? {2, 3, …, 100} 
m + n = 4, 9, 25, 49 
m + n = 4 ? 3 pairs 
m + n = 9 ? 8 pairs 
m + n = 25 ? 24 pairs 
m + n = 49 ? 48 pairs 
? 83 pairs 
? 1333 
 
 
Q7: Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let x be the number of 9-digit numbers formed using 
the digits of the set S such that only one digit is repeated and it is repeated exactly twice. 
Let y be the number of 9 -digit numbers formed using the digits of the set S such that 
only two digits are repeated and each of these is repeated exactly twice. Then,  
(a) 56x = 9y  
(b) 21x = 4y  
(c) 45x = 7y  
(d) 29x = 5y  
Page 4


JEE Main Previous Year Questions (2021-2026): 
Permutations and Combinations  
 
(January 2026) 
 
 
Q1: Three persons enter in a lift at the ground floor. The lift will go up to 10th floor. The 
number of ways, in which the three persons can exit the lift at three different floors, if the 
lift does not stop at first, second and third floors, is equal to ________. 
Ans: 210 
Sol: 
The lift goes up to the 10
th
 floor. 
The lift does not stop at 1
st
, 2
nd
, 3
rd
 floors. 
So, the persons can get down only at the remaining floors. 
4, 5, 6, 7, 8, 9, 10 ? 7 floors 
Now, all three persons must exit at three different floors (no two can get down at the same floor). 
Also, the three persons are distinct, so who gets down where matters. 
First, choose which 3 floors (out of 7) will be used: 
 
Next, assign (arrange) the 3 distinct persons to these 3 chosen floors: 
Number of arrangements of 3 persons = 3! = 6 
So, total number of ways: 
Total ways = 7C 3 × 3! = 35 × 6 = 210 
 
Q2: The number of numbers greater than 5000, less than 9000 and divisible by 3, that can 
be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is 
Ans: 42 
Sol: 
To find numbers of number greater than 5000 and less than 9000 so number will start with 5 or 
9 of the form 5abc 
and a, b, c ? {0, 1, 2, 5, 9} also the number is divisible by 3. 
This means 5 + a + b + c is divisible by 3 
so the sum a + b + c is of the type 3n + 1. 
unit place c determined by the sum of previous 3 places. 
rem(a + b) = remainder of a + b 
rem(a+b) combination (a,b) value of c total = combination (a, b) × c(choice) 
0 (0, 0), (0, 9), (9, 0), (9, 9), (1, 2), (2, 1), (5, 
1), (1, 5) 
1 8 × 1 = 8 
1 (0, 1), (9, 1), (1, 0), (1, 9), (2, 2), (2, 5), (5, 
2), (5, 5) 
0, 9 8 × 2 = 16 
2 (0, 2), (0, 5), (9, 2), (9, 5), (2, 0), (2, 9), (5, 
0), (5, 9), (1, 1) 
2, 5 9 × 2 = 18 
Total = 8 + 16 + 18 = 42 
 
 
Q3: Let S denote the set of 4-digit numbers abcd such that a > b > c > d and P denote the 
set of 5 -digit numbers having product of its digits equal to 20. Then n(S) + n(P) is equal 
to 
Ans: 260 
Sol: 
For set S: 4-digit numbers abcd with a > b > c > d 
digits are chosen from {0, 1, 2, …, 9} 
Since order is strictly fixed, any 4 distinct digits chosen will form exactly one such number. 
Number of ways to choose 4 digits from 10 is ¹ °C 4 
 
For set P: 5-digit numbers with product of digits = 20 
Possible sets of digits (using digits 1-9): 
Case 1: {5, 4, 1, 1, 1} 
Number of arrangements  
Case 2: {5, 2, 2, 1, 1} 
Number of arrangements  
n(P) = 20 + 30 = 50 
Calculating final value: 
n(S) + n(P) = 210 + 50 = 260 
The value of n(S) + n(P) is 260. 
 
 
Q4: The number of 4 -letter words, with or without meaning, which can be formed using 
the letters PQRPQRSTUVP, is _____. 
Ans: 1422 
Sol: 
Letter frequency 
P : 3 
R, Q : 2 
S, T, U, V : 1 
4 letter words can be of type 
? ABCD or AABC, AABB, AAAB 
 
= 1422 
 
 
Q5: Let ABC be a triangle. Consider four points p
1
, p
2
, p
3
, p
4
 on the side AB, five points p
5
, 
p
6
, p
7
, p
8
, p
9
 on the side BC, and four points p
10
, p
11
, p
12
, p
13
 on the side AC. None of these 
points is a vertex of the triangle ABC. Then the total number of pentagons, that can be 
formed by taking all the vertices from the points p
1
, p
2
 , …, p
13
, is 
Ans: 660 
Sol: 
4C 2 · 5C 2 · 4C 1 + 4C 2 · 5C 2 · 4C 1 + 4C 2 · 4C 2 · 5C 1 = 240 + 240 + 180 = 660 
 
Q6: Let S = {(m, n) : m, n ? {1, 2, 3, …, 50}}. If the number of elements (m, n) in S such 
that 6
m
 + 9
n
 is a multiple of 5 is p and the number of elements (m, n) in S such that m + n 
is a square of a prime number is q, then p + q is equal to _____. 
Ans: 1333 
Sol: 
5 | 6
m
 + 9
n 
? 5 | 1
m
 + (-1)
n 
? m and n has to be opposite parity. 
²C 1 × ² 5C 1 · ² 5C 1 = 625 × 2 = 1250 
For m + n = K² for some prime K. 
m + n ? {2, 3, …, 100} 
m + n = 4, 9, 25, 49 
m + n = 4 ? 3 pairs 
m + n = 9 ? 8 pairs 
m + n = 25 ? 24 pairs 
m + n = 49 ? 48 pairs 
? 83 pairs 
? 1333 
 
 
Q7: Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let x be the number of 9-digit numbers formed using 
the digits of the set S such that only one digit is repeated and it is repeated exactly twice. 
Let y be the number of 9 -digit numbers formed using the digits of the set S such that 
only two digits are repeated and each of these is repeated exactly twice. Then,  
(a) 56x = 9y  
(b) 21x = 4y  
(c) 45x = 7y  
(d) 29x = 5y  
Ans: (b) 
Sol: 
Given, S = {1, 2, 3, 4, 5, 6, 7, 8, 9} 
Calculation for x 
For x, we have to form a 9-digit number in which exactly one digit is repeated, and it is repeated 
exactly twice. So, one digit appears twice, and the remaining 7 digits appear once each. 
First, select the digit that will be repeated: from S, this can be done in ?C 1 ways. 
Now, from the remaining 8 digits, choose 7 digits that will appear once: this can be done in 8C 7 
ways. 
Now we have 9 digits in total, but one digit is repeated twice. So, the number of different 
arrangements is  ways. 
 
Calculation for y 
For y, we have to form a 9-digit number in which exactly two digits are repeated, and each of 
these is repeated exactly twice. So, two digits appear twice each (total 4 places), and the 
remaining 5 digits appear once each. 
First, select the 2 digits that will be repeated from S: this can be done in ?C 2 ways. 
Now, from the remaining 7 digits, choose 5 digits that will appear once: this can be done in 7C 5 
ways. 
Now we have 9 digits in total, with two digits repeated twice each. So, the number of different 
arrangements is = ?!/(2!2!) ways. 
 
To find the relation between x and y, divide x by y :  
To find the relation, divide x by y : 
 
So, 
 
 
Q8: The letters of the word "UDAYPUR" are written in all possible ways with or without 
meaning and these words are arranged as in a dictionary. The rank of the word 
"UDAYPUR" is 
(a) 1579 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Permutations and Combinations  
 
(January 2026) 
 
 
Q1: Three persons enter in a lift at the ground floor. The lift will go up to 10th floor. The 
number of ways, in which the three persons can exit the lift at three different floors, if the 
lift does not stop at first, second and third floors, is equal to ________. 
Ans: 210 
Sol: 
The lift goes up to the 10
th
 floor. 
The lift does not stop at 1
st
, 2
nd
, 3
rd
 floors. 
So, the persons can get down only at the remaining floors. 
4, 5, 6, 7, 8, 9, 10 ? 7 floors 
Now, all three persons must exit at three different floors (no two can get down at the same floor). 
Also, the three persons are distinct, so who gets down where matters. 
First, choose which 3 floors (out of 7) will be used: 
 
Next, assign (arrange) the 3 distinct persons to these 3 chosen floors: 
Number of arrangements of 3 persons = 3! = 6 
So, total number of ways: 
Total ways = 7C 3 × 3! = 35 × 6 = 210 
 
Q2: The number of numbers greater than 5000, less than 9000 and divisible by 3, that can 
be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is 
Ans: 42 
Sol: 
To find numbers of number greater than 5000 and less than 9000 so number will start with 5 or 
9 of the form 5abc 
and a, b, c ? {0, 1, 2, 5, 9} also the number is divisible by 3. 
This means 5 + a + b + c is divisible by 3 
so the sum a + b + c is of the type 3n + 1. 
unit place c determined by the sum of previous 3 places. 
rem(a + b) = remainder of a + b 
rem(a+b) combination (a,b) value of c total = combination (a, b) × c(choice) 
0 (0, 0), (0, 9), (9, 0), (9, 9), (1, 2), (2, 1), (5, 
1), (1, 5) 
1 8 × 1 = 8 
1 (0, 1), (9, 1), (1, 0), (1, 9), (2, 2), (2, 5), (5, 
2), (5, 5) 
0, 9 8 × 2 = 16 
2 (0, 2), (0, 5), (9, 2), (9, 5), (2, 0), (2, 9), (5, 
0), (5, 9), (1, 1) 
2, 5 9 × 2 = 18 
Total = 8 + 16 + 18 = 42 
 
 
Q3: Let S denote the set of 4-digit numbers abcd such that a > b > c > d and P denote the 
set of 5 -digit numbers having product of its digits equal to 20. Then n(S) + n(P) is equal 
to 
Ans: 260 
Sol: 
For set S: 4-digit numbers abcd with a > b > c > d 
digits are chosen from {0, 1, 2, …, 9} 
Since order is strictly fixed, any 4 distinct digits chosen will form exactly one such number. 
Number of ways to choose 4 digits from 10 is ¹ °C 4 
 
For set P: 5-digit numbers with product of digits = 20 
Possible sets of digits (using digits 1-9): 
Case 1: {5, 4, 1, 1, 1} 
Number of arrangements  
Case 2: {5, 2, 2, 1, 1} 
Number of arrangements  
n(P) = 20 + 30 = 50 
Calculating final value: 
n(S) + n(P) = 210 + 50 = 260 
The value of n(S) + n(P) is 260. 
 
 
Q4: The number of 4 -letter words, with or without meaning, which can be formed using 
the letters PQRPQRSTUVP, is _____. 
Ans: 1422 
Sol: 
Letter frequency 
P : 3 
R, Q : 2 
S, T, U, V : 1 
4 letter words can be of type 
? ABCD or AABC, AABB, AAAB 
 
= 1422 
 
 
Q5: Let ABC be a triangle. Consider four points p
1
, p
2
, p
3
, p
4
 on the side AB, five points p
5
, 
p
6
, p
7
, p
8
, p
9
 on the side BC, and four points p
10
, p
11
, p
12
, p
13
 on the side AC. None of these 
points is a vertex of the triangle ABC. Then the total number of pentagons, that can be 
formed by taking all the vertices from the points p
1
, p
2
 , …, p
13
, is 
Ans: 660 
Sol: 
4C 2 · 5C 2 · 4C 1 + 4C 2 · 5C 2 · 4C 1 + 4C 2 · 4C 2 · 5C 1 = 240 + 240 + 180 = 660 
 
Q6: Let S = {(m, n) : m, n ? {1, 2, 3, …, 50}}. If the number of elements (m, n) in S such 
that 6
m
 + 9
n
 is a multiple of 5 is p and the number of elements (m, n) in S such that m + n 
is a square of a prime number is q, then p + q is equal to _____. 
Ans: 1333 
Sol: 
5 | 6
m
 + 9
n 
? 5 | 1
m
 + (-1)
n 
? m and n has to be opposite parity. 
²C 1 × ² 5C 1 · ² 5C 1 = 625 × 2 = 1250 
For m + n = K² for some prime K. 
m + n ? {2, 3, …, 100} 
m + n = 4, 9, 25, 49 
m + n = 4 ? 3 pairs 
m + n = 9 ? 8 pairs 
m + n = 25 ? 24 pairs 
m + n = 49 ? 48 pairs 
? 83 pairs 
? 1333 
 
 
Q7: Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let x be the number of 9-digit numbers formed using 
the digits of the set S such that only one digit is repeated and it is repeated exactly twice. 
Let y be the number of 9 -digit numbers formed using the digits of the set S such that 
only two digits are repeated and each of these is repeated exactly twice. Then,  
(a) 56x = 9y  
(b) 21x = 4y  
(c) 45x = 7y  
(d) 29x = 5y  
Ans: (b) 
Sol: 
Given, S = {1, 2, 3, 4, 5, 6, 7, 8, 9} 
Calculation for x 
For x, we have to form a 9-digit number in which exactly one digit is repeated, and it is repeated 
exactly twice. So, one digit appears twice, and the remaining 7 digits appear once each. 
First, select the digit that will be repeated: from S, this can be done in ?C 1 ways. 
Now, from the remaining 8 digits, choose 7 digits that will appear once: this can be done in 8C 7 
ways. 
Now we have 9 digits in total, but one digit is repeated twice. So, the number of different 
arrangements is  ways. 
 
Calculation for y 
For y, we have to form a 9-digit number in which exactly two digits are repeated, and each of 
these is repeated exactly twice. So, two digits appear twice each (total 4 places), and the 
remaining 5 digits appear once each. 
First, select the 2 digits that will be repeated from S: this can be done in ?C 2 ways. 
Now, from the remaining 7 digits, choose 5 digits that will appear once: this can be done in 7C 5 
ways. 
Now we have 9 digits in total, with two digits repeated twice each. So, the number of different 
arrangements is = ?!/(2!2!) ways. 
 
To find the relation between x and y, divide x by y :  
To find the relation, divide x by y : 
 
So, 
 
 
Q8: The letters of the word "UDAYPUR" are written in all possible ways with or without 
meaning and these words are arranged as in a dictionary. The rank of the word 
"UDAYPUR" is 
(a) 1579 
(b) 1578 
(c) 1580 
(d) 1581 
Ans: (c) 
Sol: 
Arrang UDAYPUR in alphabetical order A, D, P, R, UU, Y 
Number of words start with A. fix A and arrange DPRUUY 
 
Similarly, number words starts from D, P, R will be same as there is arrangement of 6 letters in 
which U is repeating twice. 
So, total number of words starting with A, D, P & R are = 4 × 360 = 1440. 
Now, number of words starts with UA 
fix UA and arrange D, P, R, U, Y = 5! ways = 120. 
the next letter is D after A, so number of words starts with UDAP arranging (R, U, Y) = 3! ways = 
6 
So, number of words start with UDAR and UDAU is same = 3! + 3! = 12 
next is UDAY matches with pattern(UDAYPUR) 
number of words start with UDAYP arrang(RU) = 2! = 2 
number of words start with UDAYPR is UDAYPRU = 1 
final pattern UDAYPUR = 1 
So, rank of word UDAYPUR is = 
1440 + 120 + 6 + 12 + 1 + 1 = 1580. 
 
Q9: The largest value of n, for which 40
n
 divides 60! , is 
(a) 14 
(b) 13 
(c) 11 
(d) 12 
Ans: (a) 
Sol: 
To find largest value of n for which 60 ! is divisible by 40n 
This means we need to find exponent of 40 in 60! 
40 = 2³ × 5 
Key Concept: formula for exponent of prime p in k! is given as 
 where [] is greatest integer function. 
so, exponent of 2 in 60 !