1. In how many ways can the letters of the word PATANA can be arranged ?
Solution :
Word PATANA has 6 letters 1P, 3A, 1T, 1N
=6! / (1! 3! 1! 1!)
=120
2. How many words can be formed from the letters of the word " ENGINEERING" , so that vowels always come together ?
Solution :
Word ENGINEERING has 11 letters, from which EIEEI are vowels, they can e treated as single letter (EIEEI)NGNRNG,
So seven letters has 3 N, 2 G , 1 R and single (EIEEI)
No of arrangment
= 7! / 2!3! =240
arrangment of EIEEI
=5!/ 2!3! =10
total number of arrangemnets
= 420 * 10 = 4200 (by rule of multiplication)
3. In how many different ways can the letters of the word COMPUTER can be arranged in such a way that vowels may occupy only odd positions ?
Solution :
Here odd and even positions are :
C O M P U T E R
(O) (E) (O) (E) (O) (E) (O) (E)
Now 3 vowels O, U , E
In 5 Odd places 3 Vowels arranged as
5P3 = 5! / 2! =60
Also remaning 5 places can be arranged by C, M, P, T, R
Remaining 5 constants arranged as
5P5 =5!=120
So, required number of ways= 120 * 60 = 7200
4. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women ?
Solution:
=7C5 * 3C2
=[ 7! / (5! 2!) ]*[ 3! / (1! 2!) ]
=63
5. In a group of 5 boys and 3 girls, 3 childrens are to be selected. In how many different ways can they be selected such that at east 1 boy should be there ?
Solution :
(1boy and 2 Girls) or ( 2 boys and 1 girl ) or ( 3 boys)
=(5C1 * 3C2) + (5C2 * 3C2) + (5C3)
=55
6. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one lack ball is to be included in the draw ?
Solution :
No of ways = (drawing 1 black AND 2 others ) OR (drawing 2 black AND 1 others ) OR (drawing 3 blacks )
=(6C2 * 3C1)+(3C2 * 6C1)+(3C3)
=[ 6!/(2!4!) * 3!/(1!2!) ] +[ 3!/(2!1!) * 6!/(1!5!)] +3!/3!
=64
7. There are 5 boys and 5 girls. In how any ways they can be seated in a row so that all the girls do not sit together ?
Solution :
There are 5 boys and 5 girls, so total number of ways of sitting will be 10 ! in a row.
Now, when all girls sit together, then 5 boys and (group of 5 girls as one person ) , so total numbers will of six persons, also 5 girls can be arrenged in 5! ways,
No of ways when 5 girls sit together=6!×5!
So total no. of ways when all 5 girls do not sit together
= total number of ways of sitting 10 boys and girls - No of ways when 5 girls sit together
no of ways all 5 girls dont sit together =(10!)-(6!5!)
=3542400
8. In a party every guest shakes hand with every other guest. If there was total of 105 handshakes in the party, find the number of persons persent in the party ?
Solution :
For every handshake two persons are required,
let n be the number of persons persent in the party. So,
nC2 =105 (or)
n(n-1) /2 =105,
n^2-n-210=0
so, n=15 , -14
Total number of persons present in the party were 15
9. Five digits are given as 3, 1, 0, 9, 5
(1) From these digits how many five digits numbers can be formed, without repetition of the digits ?
(2) How many of them are divisible by 5 ?
(3) How many of them are not divisible by 5 ?
Solution :
(1) Total number of 5 digit numbers will be 5! but when 0 be at last place then it will become 4 digits so ,
Total numbers will be : 5 ! - 4 ! = 96
(2) For divisibility with 5, at unit place number should be 0 or 5
(a) when unit place has 0,(ex. 39510) then remaning 4 numbers can be arrenged in 4! = 24 ways
(b) when unit place has 5 (ex, 90135 ) then remaning 4 numbers can be arrenged in 4! = 24 ways ,
but when 0 wll be at last place (ex. 09315) then Total number of ways reduced to 4! - 3! = 18
So , Total numbers divisible by 5 will be = 24 + 18 = 42
(3) Numbers not divisible by 5 = ( Total numbers - Numbers divisible by 5 ) = 96 - 42 = 54
10. From the word MATHEMATICS
(a) How many different arrangements can be made by using all the letters in the word MATHEMATICS ?
Solution :
Word MATHEMATICS has total 11 letters out of which 2Ms, 2As, 2Ts, rest all single
=11! / (2! 2! 2!) =4989600
(b) How many of them begin with I ?
Solution :
when I will be fixed at first place , then there will be 10 letters left having 2Ms, 2As, 2Ts
=10! / (2! 2! 2!) =453600
(c) How many of them begin with M ?
Solution :
when M will be fixed at first place , then there will be 10 letters left having 2As, 2Ts
=10! / (2! 2!) =907200
11. In how many different ways can 5 persons stand in a row for a photograph?
1) 100
2) 120
3) 50
4) 5
Solution:
5! =>5*4*3*2*1
=120
12. How many different words can be formed using the letters of the word ‘BANKER’?
1) 120
2) 6
3) 720
4) 12
Solution:
6!=> 6*5*4*3*2*1 =720
13. In how many ways can the letters of the word COMPUTER be arranged?
1) 6!
2) 7!
3) 8!
4) 5040
Solution:
C O M P U T E R =8!
14. How many different 4 digit numbers can be formed using the digits 1, 2,3,6,7 and 9?
1) 120
2) 24
3) 720
4) 360
Solution:
6P4=6! / 2!
=(6*5*4*3*2*1) /2!
=720
15. How many different words can be formed using the letters of the words
(i) MIRROR (ii) BANANA (iii) SUCCESSFUL
1) 120, 60, 151200
2) 6!, 6!, 10!
3) 4!, 3!, 6!
4) 120, 120, 360
Solution:
(i) MIRROR =6! / 3! =120
(ii) BANANA = 6! / 3!2! =60
(iii) SUCCESSFUL = 10! / (3!2!2!)
=151200
16. A set of 12 books has 3 identical Quant books, 3 identical Reasoning books, 4 identical English books and 2 different books on General Awareness. In how many different ways can these 12 books be arranged in a book-shelf?
1) 12!
2) 12!/(3!x3!x4!)
3) 12!/(3!x3!x4!x2!)
4) 126
17. In how many ways can a set of chess pieces consisting of a king, a queen, two identical rooks, two identical knights and two identical bishops be placed on the first row of a chessboard?
1) 8!
2) 88
3) 5040
4) 4280
18. A father has 2 apples and 3 pears. Each weekday (Monday through Friday) he gives one of the fruits to his daughter. In how many ways can this be done?
1) 120
2) 10
3) 24
4) 12
19) 9. How many different words can be formed using the letters of the word ‘EDUCATION’ such that
(i) the word always starts with the letter ‘D’?
1) 9!
2) 8!
3) 2 x 8!
4) 8!/2
(ii) the word always ends with a vowel?
1) 5! x 8!
2) 8!
3) 5 x 8!
4) 9!
(iii) the word always begins with the letter ‘A’ and ends with a consonant?
1) 7!
2) 7! x 4!
3) 4 x 7!
4) 8! x 4
(iv) all the consonants are always together
1) 6! x 4
2) 6!
3) 2 x 8!
4) 6! x 4!
(v) the letters D, A, O and N are always together
1) 6! x 4
2) 6!
3) 2 x 8!
4) 6! x 4!
(vi) No two consonants are together
1) 6! x 6P4
2) 6!
3) 5! x 6P4
4) 6! x 2!
(vii) the letters A and T are never together
1) 7! x 8P4
2) 7!
3) 2 x 7!
4) 7! x 8P2
10. In how many ways can the letters of the word PLUMBER such that all the vowels are always together?
1) 6! x 2!
2) 7!
3) 5! x 2!
4) 6!
Solution :
Word PATANA has 6 letters 1P, 3A, 1T, 1N
=6! / (1! 3! 1! 1!)
=120
2. How many words can be formed from the letters of the word " ENGINEERING" , so that vowels always come together ?
Solution :
Word ENGINEERING has 11 letters, from which EIEEI are vowels, they can e treated as single letter (EIEEI)NGNRNG,
So seven letters has 3 N, 2 G , 1 R and single (EIEEI)
No of arrangment
= 7! / 2!3! =240
arrangment of EIEEI
=5!/ 2!3! =10
total number of arrangemnets
= 420 * 10 = 4200 (by rule of multiplication)
3. In how many different ways can the letters of the word COMPUTER can be arranged in such a way that vowels may occupy only odd positions ?
Solution :
Here odd and even positions are :
C O M P U T E R
(O) (E) (O) (E) (O) (E) (O) (E)
Now 3 vowels O, U , E
In 5 Odd places 3 Vowels arranged as
5P3 = 5! / 2! =60
Also remaning 5 places can be arranged by C, M, P, T, R
Remaining 5 constants arranged as
5P5 =5!=120
So, required number of ways= 120 * 60 = 7200
4. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women ?
Solution:
=7C5 * 3C2
=[ 7! / (5! 2!) ]*[ 3! / (1! 2!) ]
=63
5. In a group of 5 boys and 3 girls, 3 childrens are to be selected. In how many different ways can they be selected such that at east 1 boy should be there ?
Solution :
(1boy and 2 Girls) or ( 2 boys and 1 girl ) or ( 3 boys)
=(5C1 * 3C2) + (5C2 * 3C2) + (5C3)
=55
6. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one lack ball is to be included in the draw ?
Solution :
No of ways = (drawing 1 black AND 2 others ) OR (drawing 2 black AND 1 others ) OR (drawing 3 blacks )
=(6C2 * 3C1)+(3C2 * 6C1)+(3C3)
=[ 6!/(2!4!) * 3!/(1!2!) ] +[ 3!/(2!1!) * 6!/(1!5!)] +3!/3!
=64
7. There are 5 boys and 5 girls. In how any ways they can be seated in a row so that all the girls do not sit together ?
Solution :
There are 5 boys and 5 girls, so total number of ways of sitting will be 10 ! in a row.
Now, when all girls sit together, then 5 boys and (group of 5 girls as one person ) , so total numbers will of six persons, also 5 girls can be arrenged in 5! ways,
No of ways when 5 girls sit together=6!×5!
So total no. of ways when all 5 girls do not sit together
= total number of ways of sitting 10 boys and girls - No of ways when 5 girls sit together
no of ways all 5 girls dont sit together =(10!)-(6!5!)
=3542400
8. In a party every guest shakes hand with every other guest. If there was total of 105 handshakes in the party, find the number of persons persent in the party ?
Solution :
For every handshake two persons are required,
let n be the number of persons persent in the party. So,
nC2 =105 (or)
n(n-1) /2 =105,
n^2-n-210=0
so, n=15 , -14
Total number of persons present in the party were 15
9. Five digits are given as 3, 1, 0, 9, 5
(1) From these digits how many five digits numbers can be formed, without repetition of the digits ?
(2) How many of them are divisible by 5 ?
(3) How many of them are not divisible by 5 ?
Solution :
(1) Total number of 5 digit numbers will be 5! but when 0 be at last place then it will become 4 digits so ,
Total numbers will be : 5 ! - 4 ! = 96
(2) For divisibility with 5, at unit place number should be 0 or 5
(a) when unit place has 0,(ex. 39510) then remaning 4 numbers can be arrenged in 4! = 24 ways
(b) when unit place has 5 (ex, 90135 ) then remaning 4 numbers can be arrenged in 4! = 24 ways ,
but when 0 wll be at last place (ex. 09315) then Total number of ways reduced to 4! - 3! = 18
So , Total numbers divisible by 5 will be = 24 + 18 = 42
(3) Numbers not divisible by 5 = ( Total numbers - Numbers divisible by 5 ) = 96 - 42 = 54
10. From the word MATHEMATICS
(a) How many different arrangements can be made by using all the letters in the word MATHEMATICS ?
Solution :
Word MATHEMATICS has total 11 letters out of which 2Ms, 2As, 2Ts, rest all single
=11! / (2! 2! 2!) =4989600
(b) How many of them begin with I ?
Solution :
when I will be fixed at first place , then there will be 10 letters left having 2Ms, 2As, 2Ts
=10! / (2! 2! 2!) =453600
(c) How many of them begin with M ?
Solution :
when M will be fixed at first place , then there will be 10 letters left having 2As, 2Ts
=10! / (2! 2!) =907200
11. In how many different ways can 5 persons stand in a row for a photograph?
1) 100
2) 120
3) 50
4) 5
Solution:
5! =>5*4*3*2*1
=120
12. How many different words can be formed using the letters of the word ‘BANKER’?
1) 120
2) 6
3) 720
4) 12
Solution:
6!=> 6*5*4*3*2*1 =720
13. In how many ways can the letters of the word COMPUTER be arranged?
1) 6!
2) 7!
3) 8!
4) 5040
Solution:
C O M P U T E R =8!
14. How many different 4 digit numbers can be formed using the digits 1, 2,3,6,7 and 9?
1) 120
2) 24
3) 720
4) 360
Solution:
6P4=6! / 2!
=(6*5*4*3*2*1) /2!
=720
15. How many different words can be formed using the letters of the words
(i) MIRROR (ii) BANANA (iii) SUCCESSFUL
1) 120, 60, 151200
2) 6!, 6!, 10!
3) 4!, 3!, 6!
4) 120, 120, 360
Solution:
(i) MIRROR =6! / 3! =120
(ii) BANANA = 6! / 3!2! =60
(iii) SUCCESSFUL = 10! / (3!2!2!)
=151200
16. A set of 12 books has 3 identical Quant books, 3 identical Reasoning books, 4 identical English books and 2 different books on General Awareness. In how many different ways can these 12 books be arranged in a book-shelf?
1) 12!
2) 12!/(3!x3!x4!)
3) 12!/(3!x3!x4!x2!)
4) 126
17. In how many ways can a set of chess pieces consisting of a king, a queen, two identical rooks, two identical knights and two identical bishops be placed on the first row of a chessboard?
1) 8!
2) 88
3) 5040
4) 4280
18. A father has 2 apples and 3 pears. Each weekday (Monday through Friday) he gives one of the fruits to his daughter. In how many ways can this be done?
1) 120
2) 10
3) 24
4) 12
19) 9. How many different words can be formed using the letters of the word ‘EDUCATION’ such that
(i) the word always starts with the letter ‘D’?
1) 9!
2) 8!
3) 2 x 8!
4) 8!/2
(ii) the word always ends with a vowel?
1) 5! x 8!
2) 8!
3) 5 x 8!
4) 9!
(iii) the word always begins with the letter ‘A’ and ends with a consonant?
1) 7!
2) 7! x 4!
3) 4 x 7!
4) 8! x 4
(iv) all the consonants are always together
1) 6! x 4
2) 6!
3) 2 x 8!
4) 6! x 4!
(v) the letters D, A, O and N are always together
1) 6! x 4
2) 6!
3) 2 x 8!
4) 6! x 4!
(vi) No two consonants are together
1) 6! x 6P4
2) 6!
3) 5! x 6P4
4) 6! x 2!
(vii) the letters A and T are never together
1) 7! x 8P4
2) 7!
3) 2 x 7!
4) 7! x 8P2
10. In how many ways can the letters of the word PLUMBER such that all the vowels are always together?
1) 6! x 2!
2) 7!
3) 5! x 2!
4) 6!
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