Permutation & Combination Solved Examples | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


JEE Solved Examples on 
Permutation & Combination 
Q:1 In a monthly test, the teacher decides that there will be three questions, one from 
each of exercise 7 , 8 and 9 of the text book. If there are 12 questions in exercise 7,18 
in exercise 8 and 9 in exercise 9 , in how many ways can three questions be selected 
(a) 1944 
(b) 1499 
(c) 4991 
(d) None of these 
Solution: (a) There are 12 questions in exercise 7. So, one question from exercise 7 can be 
selected in 12 ways. Exercise 8 contains 18 questions. So, second question can be selected 
in 18 ways. There are 9 questions in exercise 9. So, third question can be selected in 9 
ways. Hence, three questions can be selected in 12 × 18 × 9 = 1944 ways. 
Q: 2 How many numbers can be made with the help of the digits ?? , ?? , ?? , ?? , ?? , ?? which are 
greater than 3000 (repetition is not allowed) 
(a) 180 
(b) 360 
(c) 1380 
(d) 1500 
Solution: (c) All the 5 digit numbers and 6 digit numbers are greater than 3000. Therefore 
number of 5 digit numbers  = 
6
?? 5
- 
5
?? 5
= 600. 
{Since the case that o will be at ten thousand place should be omit}. Similarly number of 6 
digit numbers 6! - 5! = 600. 
Q: 3 The number of 5 digit telephone numbers having at least one of their digits 
repeated is 
(a) 90,000 
(b) 100,000 
(c) 30,240 
(d) 69,760 
Solution: (d) Using the digits 0,1,2, … . . . ,9 the number of five digit telephone numbers which 
can be formed is 10
5
. (since repetition is allowed) 
The number of five digit telephone numbers which have none of the digits repeated =
 
10
?? 5
= 30240 
? The required number of telephone numbers = 10
5
- 30240= 69760. 
Q: 4 If the letters of the word 'KRISNA' are arranged in all possible ways and these 
words are written out as in a dictionary, then the rank of the word 'KRISNA' is 
(a) 324 
(b) 341 
(c) 359 
(d) None of these 
Page 2


JEE Solved Examples on 
Permutation & Combination 
Q:1 In a monthly test, the teacher decides that there will be three questions, one from 
each of exercise 7 , 8 and 9 of the text book. If there are 12 questions in exercise 7,18 
in exercise 8 and 9 in exercise 9 , in how many ways can three questions be selected 
(a) 1944 
(b) 1499 
(c) 4991 
(d) None of these 
Solution: (a) There are 12 questions in exercise 7. So, one question from exercise 7 can be 
selected in 12 ways. Exercise 8 contains 18 questions. So, second question can be selected 
in 18 ways. There are 9 questions in exercise 9. So, third question can be selected in 9 
ways. Hence, three questions can be selected in 12 × 18 × 9 = 1944 ways. 
Q: 2 How many numbers can be made with the help of the digits ?? , ?? , ?? , ?? , ?? , ?? which are 
greater than 3000 (repetition is not allowed) 
(a) 180 
(b) 360 
(c) 1380 
(d) 1500 
Solution: (c) All the 5 digit numbers and 6 digit numbers are greater than 3000. Therefore 
number of 5 digit numbers  = 
6
?? 5
- 
5
?? 5
= 600. 
{Since the case that o will be at ten thousand place should be omit}. Similarly number of 6 
digit numbers 6! - 5! = 600. 
Q: 3 The number of 5 digit telephone numbers having at least one of their digits 
repeated is 
(a) 90,000 
(b) 100,000 
(c) 30,240 
(d) 69,760 
Solution: (d) Using the digits 0,1,2, … . . . ,9 the number of five digit telephone numbers which 
can be formed is 10
5
. (since repetition is allowed) 
The number of five digit telephone numbers which have none of the digits repeated =
 
10
?? 5
= 30240 
? The required number of telephone numbers = 10
5
- 30240= 69760. 
Q: 4 If the letters of the word 'KRISNA' are arranged in all possible ways and these 
words are written out as in a dictionary, then the rank of the word 'KRISNA' is 
(a) 324 
(b) 341 
(c) 359 
(d) None of these 
Solution: (a) 
Words starting from ?? are 5! = 120; 
Words starting from ?? are 5! = 120 
Words starting from ???? are 4! = 24; 
Words starting from ???? are 4! = 24; 
Words starting from ???? are 4! = 24 
Words starting from KRIA are 2! = 2; 
Words starting from ?????? are 3! = 6 
Words starting from KRIS are 1! = 1 
Words starting from KRIN are 2! = 2 
Hence rank of the word KRISNA is 324 
Words starting from KRISNA are 1! = 1 
Q: 5 The number of numbers that can be formed with the help of the digits 1, 2, 3, 4, 3, 
2, 1 so that odd digits always occupy odd places, is 
(a) 24 
(b) 18 
(c) 12 
(d) 30 
Solution: (b) The 4 odd digits 1, 3, 3, 1 can be arranged in the 4 odd places, in 
4!
2!2!
= 6 ways 
and 3 even digits 2,4,2 can be arranged in the three even places 
3!
2!
= 3 ways. Hence the 
required number of ways = 6 × 3 = 18. 
Q6. In how many ways can 5 boys and 5 girls sit in a circle so that no boys sit 
together 
(a) ?? ! × ?? ! 
(b) ?? ! × ?? ! 
(c) 
?? !×?? !
?? 
(d) None of these 
Solution: (b) Since total number of ways in which boys can occupy any place is (5 - 1)! = 4 
! and the 5 girls can be sit accordingly in 5 ! ways. Hence required number of ways are 4! × 5 
!. 
Q7.  The number of ways in which 5 beads of different colours form a necklace is 
(a) 12 
(b) 24 
(c) 120 
(d) 60 
Solution: (a) The number of ways in which 5 beads of different colours can be arranged in a 
circle to form a necklace are 
= (5 - 1)! = 4 !.  
But the clockwise and anticlockwise arrangement are not different (because when the 
necklace is turned over one gives rise to another). Hence the total number of ways of 
arranging the beads = 
1
2
(4!) = 12. 
Page 3


JEE Solved Examples on 
Permutation & Combination 
Q:1 In a monthly test, the teacher decides that there will be three questions, one from 
each of exercise 7 , 8 and 9 of the text book. If there are 12 questions in exercise 7,18 
in exercise 8 and 9 in exercise 9 , in how many ways can three questions be selected 
(a) 1944 
(b) 1499 
(c) 4991 
(d) None of these 
Solution: (a) There are 12 questions in exercise 7. So, one question from exercise 7 can be 
selected in 12 ways. Exercise 8 contains 18 questions. So, second question can be selected 
in 18 ways. There are 9 questions in exercise 9. So, third question can be selected in 9 
ways. Hence, three questions can be selected in 12 × 18 × 9 = 1944 ways. 
Q: 2 How many numbers can be made with the help of the digits ?? , ?? , ?? , ?? , ?? , ?? which are 
greater than 3000 (repetition is not allowed) 
(a) 180 
(b) 360 
(c) 1380 
(d) 1500 
Solution: (c) All the 5 digit numbers and 6 digit numbers are greater than 3000. Therefore 
number of 5 digit numbers  = 
6
?? 5
- 
5
?? 5
= 600. 
{Since the case that o will be at ten thousand place should be omit}. Similarly number of 6 
digit numbers 6! - 5! = 600. 
Q: 3 The number of 5 digit telephone numbers having at least one of their digits 
repeated is 
(a) 90,000 
(b) 100,000 
(c) 30,240 
(d) 69,760 
Solution: (d) Using the digits 0,1,2, … . . . ,9 the number of five digit telephone numbers which 
can be formed is 10
5
. (since repetition is allowed) 
The number of five digit telephone numbers which have none of the digits repeated =
 
10
?? 5
= 30240 
? The required number of telephone numbers = 10
5
- 30240= 69760. 
Q: 4 If the letters of the word 'KRISNA' are arranged in all possible ways and these 
words are written out as in a dictionary, then the rank of the word 'KRISNA' is 
(a) 324 
(b) 341 
(c) 359 
(d) None of these 
Solution: (a) 
Words starting from ?? are 5! = 120; 
Words starting from ?? are 5! = 120 
Words starting from ???? are 4! = 24; 
Words starting from ???? are 4! = 24; 
Words starting from ???? are 4! = 24 
Words starting from KRIA are 2! = 2; 
Words starting from ?????? are 3! = 6 
Words starting from KRIS are 1! = 1 
Words starting from KRIN are 2! = 2 
Hence rank of the word KRISNA is 324 
Words starting from KRISNA are 1! = 1 
Q: 5 The number of numbers that can be formed with the help of the digits 1, 2, 3, 4, 3, 
2, 1 so that odd digits always occupy odd places, is 
(a) 24 
(b) 18 
(c) 12 
(d) 30 
Solution: (b) The 4 odd digits 1, 3, 3, 1 can be arranged in the 4 odd places, in 
4!
2!2!
= 6 ways 
and 3 even digits 2,4,2 can be arranged in the three even places 
3!
2!
= 3 ways. Hence the 
required number of ways = 6 × 3 = 18. 
Q6. In how many ways can 5 boys and 5 girls sit in a circle so that no boys sit 
together 
(a) ?? ! × ?? ! 
(b) ?? ! × ?? ! 
(c) 
?? !×?? !
?? 
(d) None of these 
Solution: (b) Since total number of ways in which boys can occupy any place is (5 - 1)! = 4 
! and the 5 girls can be sit accordingly in 5 ! ways. Hence required number of ways are 4! × 5 
!. 
Q7.  The number of ways in which 5 beads of different colours form a necklace is 
(a) 12 
(b) 24 
(c) 120 
(d) 60 
Solution: (a) The number of ways in which 5 beads of different colours can be arranged in a 
circle to form a necklace are 
= (5 - 1)! = 4 !.  
But the clockwise and anticlockwise arrangement are not different (because when the 
necklace is turned over one gives rise to another). Hence the total number of ways of 
arranging the beads = 
1
2
(4!) = 12. 
Q8.  
 
?? ?? ?? 
?? ?? ?? -?? = 
(a) 
?? -?? ?? 
(b) 
?? +?? -?? ?? 
(c) 
?? -?? +?? ?? 
(d) 
?? -?? -?? ?? 
Solution: (c)  
 
?? ?? ?? 
?? ?? ?? -1
=
?? !
?? !(?? -?? )!
(?? -1)!(?? -?? +1)!
?
?? !
?? !(?? -?? )!
×
(?? -1)!(?? -?? +1)!
?? !
=
(?? -?? +1)(?? -1)!(?? -?? )!
?? (?? -1)!(?? -?? )!
=
(?? -?? +1)
?? . 
Q9.  In a conference of 8 persons, if each person shake hand with the other one only, 
then the total number of shake hands shall be 
(a) 64 
(b) 56 
(c) 49 
(d) 28 
Solution: (d) Total number of shake hands when each person shake hands with the other 
once only = 
8
?? 2
= 28 ways. 
Q10. 10 different letters of English alphabet are given. Out of these letters, words of 5 
letters are formed. How many words are formed when atleast one letter is repeated 
(a) 99748 
(b) 98748 
(c) 96747 
(d) 97147 
Solution: (a) Number of words of 5 letters in which letters have been repeated any times =
10
5
 
But number of words on taking 5 different letters out of 10 = 
10
?? 5
= 252 
? Required number of words = 10
5
- 252 = 99748. 
Q11.  In the 13 cricket players 4 are bowlers, then how many ways can form a cricket 
team of 11 players in which at least 
2 bowlers included 
(a) 55 
(b) 72 
(c) 78 
(d) None of these 
Solution: (c) The number of ways can be given as follows: 
2 bowlers and 9 other players = 
4
?? 2
× 
9
?? 9
; 3 bowlers and 8 other players = 
4
?? 3
× 
9
?? 8
4 bowlers and 7 other players = 
4
?? 4
× 
9
?? 7
 Hence required number of ways = 6 × 1 + 4 × 9 + 1 × 36 = 78.
 
 
 
Page 4


JEE Solved Examples on 
Permutation & Combination 
Q:1 In a monthly test, the teacher decides that there will be three questions, one from 
each of exercise 7 , 8 and 9 of the text book. If there are 12 questions in exercise 7,18 
in exercise 8 and 9 in exercise 9 , in how many ways can three questions be selected 
(a) 1944 
(b) 1499 
(c) 4991 
(d) None of these 
Solution: (a) There are 12 questions in exercise 7. So, one question from exercise 7 can be 
selected in 12 ways. Exercise 8 contains 18 questions. So, second question can be selected 
in 18 ways. There are 9 questions in exercise 9. So, third question can be selected in 9 
ways. Hence, three questions can be selected in 12 × 18 × 9 = 1944 ways. 
Q: 2 How many numbers can be made with the help of the digits ?? , ?? , ?? , ?? , ?? , ?? which are 
greater than 3000 (repetition is not allowed) 
(a) 180 
(b) 360 
(c) 1380 
(d) 1500 
Solution: (c) All the 5 digit numbers and 6 digit numbers are greater than 3000. Therefore 
number of 5 digit numbers  = 
6
?? 5
- 
5
?? 5
= 600. 
{Since the case that o will be at ten thousand place should be omit}. Similarly number of 6 
digit numbers 6! - 5! = 600. 
Q: 3 The number of 5 digit telephone numbers having at least one of their digits 
repeated is 
(a) 90,000 
(b) 100,000 
(c) 30,240 
(d) 69,760 
Solution: (d) Using the digits 0,1,2, … . . . ,9 the number of five digit telephone numbers which 
can be formed is 10
5
. (since repetition is allowed) 
The number of five digit telephone numbers which have none of the digits repeated =
 
10
?? 5
= 30240 
? The required number of telephone numbers = 10
5
- 30240= 69760. 
Q: 4 If the letters of the word 'KRISNA' are arranged in all possible ways and these 
words are written out as in a dictionary, then the rank of the word 'KRISNA' is 
(a) 324 
(b) 341 
(c) 359 
(d) None of these 
Solution: (a) 
Words starting from ?? are 5! = 120; 
Words starting from ?? are 5! = 120 
Words starting from ???? are 4! = 24; 
Words starting from ???? are 4! = 24; 
Words starting from ???? are 4! = 24 
Words starting from KRIA are 2! = 2; 
Words starting from ?????? are 3! = 6 
Words starting from KRIS are 1! = 1 
Words starting from KRIN are 2! = 2 
Hence rank of the word KRISNA is 324 
Words starting from KRISNA are 1! = 1 
Q: 5 The number of numbers that can be formed with the help of the digits 1, 2, 3, 4, 3, 
2, 1 so that odd digits always occupy odd places, is 
(a) 24 
(b) 18 
(c) 12 
(d) 30 
Solution: (b) The 4 odd digits 1, 3, 3, 1 can be arranged in the 4 odd places, in 
4!
2!2!
= 6 ways 
and 3 even digits 2,4,2 can be arranged in the three even places 
3!
2!
= 3 ways. Hence the 
required number of ways = 6 × 3 = 18. 
Q6. In how many ways can 5 boys and 5 girls sit in a circle so that no boys sit 
together 
(a) ?? ! × ?? ! 
(b) ?? ! × ?? ! 
(c) 
?? !×?? !
?? 
(d) None of these 
Solution: (b) Since total number of ways in which boys can occupy any place is (5 - 1)! = 4 
! and the 5 girls can be sit accordingly in 5 ! ways. Hence required number of ways are 4! × 5 
!. 
Q7.  The number of ways in which 5 beads of different colours form a necklace is 
(a) 12 
(b) 24 
(c) 120 
(d) 60 
Solution: (a) The number of ways in which 5 beads of different colours can be arranged in a 
circle to form a necklace are 
= (5 - 1)! = 4 !.  
But the clockwise and anticlockwise arrangement are not different (because when the 
necklace is turned over one gives rise to another). Hence the total number of ways of 
arranging the beads = 
1
2
(4!) = 12. 
Q8.  
 
?? ?? ?? 
?? ?? ?? -?? = 
(a) 
?? -?? ?? 
(b) 
?? +?? -?? ?? 
(c) 
?? -?? +?? ?? 
(d) 
?? -?? -?? ?? 
Solution: (c)  
 
?? ?? ?? 
?? ?? ?? -1
=
?? !
?? !(?? -?? )!
(?? -1)!(?? -?? +1)!
?
?? !
?? !(?? -?? )!
×
(?? -1)!(?? -?? +1)!
?? !
=
(?? -?? +1)(?? -1)!(?? -?? )!
?? (?? -1)!(?? -?? )!
=
(?? -?? +1)
?? . 
Q9.  In a conference of 8 persons, if each person shake hand with the other one only, 
then the total number of shake hands shall be 
(a) 64 
(b) 56 
(c) 49 
(d) 28 
Solution: (d) Total number of shake hands when each person shake hands with the other 
once only = 
8
?? 2
= 28 ways. 
Q10. 10 different letters of English alphabet are given. Out of these letters, words of 5 
letters are formed. How many words are formed when atleast one letter is repeated 
(a) 99748 
(b) 98748 
(c) 96747 
(d) 97147 
Solution: (a) Number of words of 5 letters in which letters have been repeated any times =
10
5
 
But number of words on taking 5 different letters out of 10 = 
10
?? 5
= 252 
? Required number of words = 10
5
- 252 = 99748. 
Q11.  In the 13 cricket players 4 are bowlers, then how many ways can form a cricket 
team of 11 players in which at least 
2 bowlers included 
(a) 55 
(b) 72 
(c) 78 
(d) None of these 
Solution: (c) The number of ways can be given as follows: 
2 bowlers and 9 other players = 
4
?? 2
× 
9
?? 9
; 3 bowlers and 8 other players = 
4
?? 3
× 
9
?? 8
4 bowlers and 7 other players = 
4
?? 4
× 
9
?? 7
 Hence required number of ways = 6 × 1 + 4 × 9 + 1 × 36 = 78.
 
 
 
Q12. In how many ways can 6 persons to be selected from 4 officers and 8 
constables, if at least one officer is to be included 
 
(a) 224 
(b) 672 
(c) 896 
(d) None of these 
Solution: (c) Required number of ways = 
4
?? 1
× 
8
?? 5
+ 
4
?? 2
× 
8
?? 4
+ 
4
?? 3
× 
8
?? 3
+
 
4
?? 4
× 
8
?? 2
= 4 × 56 + 6 × 70 + 4 × 56 + 1 × 28 = 896. 
 
Q13. A question paper is divided into two parts ?? and ?? and each part contains 5 
questions. The number of ways in which a candidate can answer 6 questions 
selecting at least two questions from each part is 
(a) 80 
(b) 100 
(c) 200 
(d) None of these 
Solution: (c) The number of ways that the candidate may select 
2 questions from ?? and 4 from ?? = 
5
?? 2
× 
5
?? 4
; 3 questions form ?? and 3 from ?? = 
5
?? 3
× 
5
?? 3
 
4 questions from ?? and 2 from ?? = 
5
?? 4
× 
5
?? 2
. Hence total number of ways are 200. 
Q14.  The number of straight lines joining 8 points on a circle is 
(a) 8 
(b) 16 
(c) 24 
(d) 28 
Solution: (d) Number of straight line = 
8
?? 2
= 28. 
 
 
 
 
 
 
 
 
 
Page 5


JEE Solved Examples on 
Permutation & Combination 
Q:1 In a monthly test, the teacher decides that there will be three questions, one from 
each of exercise 7 , 8 and 9 of the text book. If there are 12 questions in exercise 7,18 
in exercise 8 and 9 in exercise 9 , in how many ways can three questions be selected 
(a) 1944 
(b) 1499 
(c) 4991 
(d) None of these 
Solution: (a) There are 12 questions in exercise 7. So, one question from exercise 7 can be 
selected in 12 ways. Exercise 8 contains 18 questions. So, second question can be selected 
in 18 ways. There are 9 questions in exercise 9. So, third question can be selected in 9 
ways. Hence, three questions can be selected in 12 × 18 × 9 = 1944 ways. 
Q: 2 How many numbers can be made with the help of the digits ?? , ?? , ?? , ?? , ?? , ?? which are 
greater than 3000 (repetition is not allowed) 
(a) 180 
(b) 360 
(c) 1380 
(d) 1500 
Solution: (c) All the 5 digit numbers and 6 digit numbers are greater than 3000. Therefore 
number of 5 digit numbers  = 
6
?? 5
- 
5
?? 5
= 600. 
{Since the case that o will be at ten thousand place should be omit}. Similarly number of 6 
digit numbers 6! - 5! = 600. 
Q: 3 The number of 5 digit telephone numbers having at least one of their digits 
repeated is 
(a) 90,000 
(b) 100,000 
(c) 30,240 
(d) 69,760 
Solution: (d) Using the digits 0,1,2, … . . . ,9 the number of five digit telephone numbers which 
can be formed is 10
5
. (since repetition is allowed) 
The number of five digit telephone numbers which have none of the digits repeated =
 
10
?? 5
= 30240 
? The required number of telephone numbers = 10
5
- 30240= 69760. 
Q: 4 If the letters of the word 'KRISNA' are arranged in all possible ways and these 
words are written out as in a dictionary, then the rank of the word 'KRISNA' is 
(a) 324 
(b) 341 
(c) 359 
(d) None of these 
Solution: (a) 
Words starting from ?? are 5! = 120; 
Words starting from ?? are 5! = 120 
Words starting from ???? are 4! = 24; 
Words starting from ???? are 4! = 24; 
Words starting from ???? are 4! = 24 
Words starting from KRIA are 2! = 2; 
Words starting from ?????? are 3! = 6 
Words starting from KRIS are 1! = 1 
Words starting from KRIN are 2! = 2 
Hence rank of the word KRISNA is 324 
Words starting from KRISNA are 1! = 1 
Q: 5 The number of numbers that can be formed with the help of the digits 1, 2, 3, 4, 3, 
2, 1 so that odd digits always occupy odd places, is 
(a) 24 
(b) 18 
(c) 12 
(d) 30 
Solution: (b) The 4 odd digits 1, 3, 3, 1 can be arranged in the 4 odd places, in 
4!
2!2!
= 6 ways 
and 3 even digits 2,4,2 can be arranged in the three even places 
3!
2!
= 3 ways. Hence the 
required number of ways = 6 × 3 = 18. 
Q6. In how many ways can 5 boys and 5 girls sit in a circle so that no boys sit 
together 
(a) ?? ! × ?? ! 
(b) ?? ! × ?? ! 
(c) 
?? !×?? !
?? 
(d) None of these 
Solution: (b) Since total number of ways in which boys can occupy any place is (5 - 1)! = 4 
! and the 5 girls can be sit accordingly in 5 ! ways. Hence required number of ways are 4! × 5 
!. 
Q7.  The number of ways in which 5 beads of different colours form a necklace is 
(a) 12 
(b) 24 
(c) 120 
(d) 60 
Solution: (a) The number of ways in which 5 beads of different colours can be arranged in a 
circle to form a necklace are 
= (5 - 1)! = 4 !.  
But the clockwise and anticlockwise arrangement are not different (because when the 
necklace is turned over one gives rise to another). Hence the total number of ways of 
arranging the beads = 
1
2
(4!) = 12. 
Q8.  
 
?? ?? ?? 
?? ?? ?? -?? = 
(a) 
?? -?? ?? 
(b) 
?? +?? -?? ?? 
(c) 
?? -?? +?? ?? 
(d) 
?? -?? -?? ?? 
Solution: (c)  
 
?? ?? ?? 
?? ?? ?? -1
=
?? !
?? !(?? -?? )!
(?? -1)!(?? -?? +1)!
?
?? !
?? !(?? -?? )!
×
(?? -1)!(?? -?? +1)!
?? !
=
(?? -?? +1)(?? -1)!(?? -?? )!
?? (?? -1)!(?? -?? )!
=
(?? -?? +1)
?? . 
Q9.  In a conference of 8 persons, if each person shake hand with the other one only, 
then the total number of shake hands shall be 
(a) 64 
(b) 56 
(c) 49 
(d) 28 
Solution: (d) Total number of shake hands when each person shake hands with the other 
once only = 
8
?? 2
= 28 ways. 
Q10. 10 different letters of English alphabet are given. Out of these letters, words of 5 
letters are formed. How many words are formed when atleast one letter is repeated 
(a) 99748 
(b) 98748 
(c) 96747 
(d) 97147 
Solution: (a) Number of words of 5 letters in which letters have been repeated any times =
10
5
 
But number of words on taking 5 different letters out of 10 = 
10
?? 5
= 252 
? Required number of words = 10
5
- 252 = 99748. 
Q11.  In the 13 cricket players 4 are bowlers, then how many ways can form a cricket 
team of 11 players in which at least 
2 bowlers included 
(a) 55 
(b) 72 
(c) 78 
(d) None of these 
Solution: (c) The number of ways can be given as follows: 
2 bowlers and 9 other players = 
4
?? 2
× 
9
?? 9
; 3 bowlers and 8 other players = 
4
?? 3
× 
9
?? 8
4 bowlers and 7 other players = 
4
?? 4
× 
9
?? 7
 Hence required number of ways = 6 × 1 + 4 × 9 + 1 × 36 = 78.
 
 
 
Q12. In how many ways can 6 persons to be selected from 4 officers and 8 
constables, if at least one officer is to be included 
 
(a) 224 
(b) 672 
(c) 896 
(d) None of these 
Solution: (c) Required number of ways = 
4
?? 1
× 
8
?? 5
+ 
4
?? 2
× 
8
?? 4
+ 
4
?? 3
× 
8
?? 3
+
 
4
?? 4
× 
8
?? 2
= 4 × 56 + 6 × 70 + 4 × 56 + 1 × 28 = 896. 
 
Q13. A question paper is divided into two parts ?? and ?? and each part contains 5 
questions. The number of ways in which a candidate can answer 6 questions 
selecting at least two questions from each part is 
(a) 80 
(b) 100 
(c) 200 
(d) None of these 
Solution: (c) The number of ways that the candidate may select 
2 questions from ?? and 4 from ?? = 
5
?? 2
× 
5
?? 4
; 3 questions form ?? and 3 from ?? = 
5
?? 3
× 
5
?? 3
 
4 questions from ?? and 2 from ?? = 
5
?? 4
× 
5
?? 2
. Hence total number of ways are 200. 
Q14.  The number of straight lines joining 8 points on a circle is 
(a) 8 
(b) 16 
(c) 24 
(d) 28 
Solution: (d) Number of straight line = 
8
?? 2
= 28. 
 
 
 
 
 
 
 
 
 
Q15.  If ?? , ?? and ?? are positive integers, then  
?? ?? ?? + 
?? ?? ?? -?? ?? ?? ?? + 
?? ?? ?? -?? 
?? ?? ?? + ? . . + 
?? ?? ?? = 
 
(a) 
?? !?? !
?? !
 
(b) 
(?? +?? )!
?? !
 
(c)  
?? +?? ?? ?? 
(d)  
????
?? ?? 
Solution: (c) The result  
?? +?? ?? ?? is trivially true for ?? = 1,2 it can be easily proved by the 
principle of mathematical induction that the result is true for ?? also. 
 
 
Q16.  The number of ways dividing 52 cards amongst four players equally, are 
(a) 
???? !
(???? !)
?? 
(b) 
???? !
(???? !)
?? ?? !
 
(c) 
???? !
(???? !)
?? ?? !
 
(d) None of these 
Solution: (a) Required number of ways = 
52
?? 13
× 
39
?? 13
× 
26
?? 13
× 
13
?? 13
=
52!
39!13!
×
39!
26!13!
×
26!
13!13!
×
13!
13!
=
52!
(13!)
4
. 
 
Q17. Let ?? be the set of 4 -digit numbers ?? ?? ?? ?? ?? ?? ?? ?? where ?? ?? > ?? ?? > ?? ?? > ?? ?? , then ?? (?? ) 
is equal to 
(A) 126 
(B) 84 
(C) 210 
(D) none of these 
 
Sol : (C) 
Any selection of four digits from the ten digits 0,1,2,3, … 9 gives one such number. SO, the 
required number of 
 numbers = 10C4
 = 210.