Permutation level two

C.S.VEERARAGAVAN
98948 34264
veeraa1729@gmail.com
PERMUTATION LEVEL TWO

PERMUTATION
12 June 2015C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 2
When a dice is rolled for n times, then the number of total
out comes as
(1) 6 (2) 6n-1(3) 6n (4)6n+1
The number of total out comes is 6n
01

PERMUTATION
12 June 2015
C.S.VEERARAGAVAN 9894834264
VEERAA1729@GMAIL.COM
3
When a coin is tossed for (n-1) times, then the
number of total out comes is
(1) 2 (2) 2n (3) 2n+1 (4)2n-1
the number of total out comes is 2n-1
02

PERMUTATION
12 June 2015
4
Find the value of 35P34
(1) 35! (2) 35
(3) 34! (4) 34
05

PERMUTATION
12 June 2015
5
If nCr = nPr then the value of r can be
(1) 0 (2) 1
(3) 2 (4) More than one of the above
nPr = nCr .* r!
Hence r could be zero or one.
07
(4) More than one of the above

PERMUTATION
12 June 2015
6
The number of ways of arranging 6 persons in a
row is
(1) 6! (2) (-1)! (3) 6 (4)
08

PERMUTATION
12 June 2015
7
How many words can be formed using all the letters of
the word GINGER?
(1) 720 (2) 240 (3) 380 (4)360
Number of letters = 6
G repeated two times.
Hence number of words =
6!
2!
=3 x 4 x 5 x 6 =360
09

PERMUTATION
12 June 2015
8
How many different words, which begin with N can be
formed using all the letters of the word COUNTRY?
(1) 720 (2) 120 (3) 24 (4)5040
The word COUNTRY has 7 letters.
Since the first letter is N, the remaining 6 places can be
filled with 6 letters in 6! Ways. = 720.
11

PERMUTATION
12 June 2015
9
How many three-digit numbers can be formed using the digits
{1,2,3,4,5} , so that each digit is repeated any number of times?
(1) 150 (2) 200 (3) 25 (4)125
Consider 3 blanks
Since there are 5 digits, each blank can be filled in 5 ways.
Total number of ways is (5) (5) (5) = 125.
14

PERMUTATION
12 June 2015
10
All possible four-digit numbers, with distinct digits are
formed, using the digits {1,3,4,5,6}.
How many of them are divisible by 5?
(1) 8 (2) 12 (3) 24 (4)20
Consider four blanks
The units place is filled with 5.
The remaining three blanks can be filled with 4 digits in 4P3
ways.
The number of four digit numbers required is 4 (3) (2) or 24.
16

PERMUTATION
12 June 2015
11
In how many ways can 3 boys and 2 girls be seated in arrow,
so that all girls sit together?
(1) 12 (2) 24 (3) 84 (4)48
Let All the girls be one unit.
There are 3 boys and 1 unit of girls arranged in 4! Ways.
The two girls can be arranged among themselves in 2! Ways.
Total number of arrangements 4! 2! = 24 (2) = 48.
17

PERMUTATION
12 June 2015
12
In how many ways can 4 boys be seated in 6 chairs ?
(1)180 (2) 720 (3) 360 (4)240
4 boys be seated in 6 chairs in 6P4 = 360 ways.
22

PERMUTATION WITH RESTRICTIONS
12 June 2015
13
A basket contains 3 bananas and 4 apples. In how many
ways can one select one or more fruits?
(1) 17 (2)20 (3) 18 (4)19
Bananas can be selected in 4 ways and
Apples can be selected in 5 ways
Number of ways of selecting fruits is 4 x 5 = 20.
Number of ways of selecting one or more = 20 – 1 = 19.
30

PERMUTATION WITH RESTRICTIONS
12 June 2015
14
How many different words can be formed using the letters
of the word MARKET so that they begin with K and end
with R?
(1) 42 (2) 24 (3) 12 (4)64
Since the first place and the last place are to be filled with
K and R, the remaining four places can be arranged
In 4! Or 24 ways.
26

COMBINATION
12 June 2015
15
When a coin is tossed for n times then the number
of ways of getting exactly ‘r’ heads is
(1) 2r (2) nCr (3) nPr (4)2n
03

COMBINATION
12 June 2015
16
Find the value of 120C119
(1) 119! (2) 120! (3) 120 (4)119
04

COMBINATION
12 June 2015
17
If nC5=nC7, then 2n+1C2 is
(1) 250 (2) 300 (3) 240 (4)280
nCr=nCs  r + s = n
5 + 7 = 12
25C2 =
25 X 24
2
= 300
06

COMBINATION
12 June 2015
18
The number of ways of selecting 4 members from a
group of 10 members so that one particular member is
always included is
(1) 63 (2) 72 (3) 84 (4)56
Since one particular member is always included, we have
to select 3 members from 9 members.
This can be done in 9C3 = 84 ways.
10

COMBINATION
12 June 2015
19
In how many ways can two consonants be selected from
the English alphabet?
(1) 420 (2) 105 (3) 210 (4)300
There are 21 consonants.
Two consonants can be selected from 21 consonants in
21C2 = 210
12

COMBINATION
12 June 2015
20
A bag contains 3 white balls, 4 green balls and 5 red balls.
In how many ways two balls can be selected?
(1) 132 (2) 66 (3) 33 (4)76
Total balls = 12.
12C2 = 66 ways.
13

COMBINATION
12 June 2015
21
Ten points are selected on a plane, such that no three of
them are collinear. How many different straight lines can
be formed by joining these points?
(1) 54 (2) 45 (3) 90 (4)108
A straight line is formed by joining any two points.
Two points can be selected from 10 points in 10C2 ways.
i.e., 45 ways.
18

COMBINATION
12 June 2015
22
If the number of diagonals of a polygon is 5 times the number
of sides, the polygon is
(1) 13 (2) 20 (3) 15 (4)17
Let the number of sides be n.
Number of diagonals =
n(n−3)
2
Given
n(n−3)
2
= 5n,
N-3 = 10
N = 13
19

COMBINATION
12 June 2015
23
Rahul has 6 friends. In how many ways can he
invite 5 or more friends for dinner?
(1) 1 (2) 6 (3) 7 (4)8
5 friends can be invited in 6C5 ways.
6 friends can be invited in 6C6 ways.
Total = 6C5 + 6C6 = 6 + 1 = 7
29

COMBINATION
12 June 2015
24
In how many ways can four letters be selected from the
word EDUCATION ?
(1)9C8 (2)9C7 (3) 9C4 (4) 9C6
The number of letters in the word EDUCATION is 9.
4 letters can be selected from these 9 letters in 9C4 ways.
24

COMBINATION
12 June 2015
25
In how many ways can 3 blue balls be selected from a
bag which contains 4 white balls and 6 blue balls ?
(1) 20 (2) 10 (3) 120 (4)210
The number of blue balls is 6.
3 balls can be selected from 6 blue balls in 6C3 ways = 20
27

CIRCULAR PERMUTATION
12 June 2015
26
In how many ways can 5 men and 3 women be seated
around a circular table?
(1) 720 (2) 5040 (3) 4020 (4)2520
N persons can be positioned around a circle in (n-1)!
ways.
Hence 8 persons can be arranged in 7! = 5040 ways.
20

12 June 2015
27
How many odd numbers can be formed using the digits
{0,2,4,6} ?
(1)0 (2) 192 (3) 18 (4)20
Since all given numbers are even, we could not get any
odd number.
15

CIRCULAR PERMUTATION
12 June 2015
29
If n books can be arranged on an ordinary shelf in 720 ways,
then in how many ways can these books be arranged on a
circular shelf ?
(1)120 (2) 720 (3) 360 (4)60
N books can be arranged in n! ways.
Given n! = 720 = 6!
N = 6
6 books can be arranged on a circular shelf in (6-1)! =5!
120 ways.
23

COMPOUND EVENT
12 June 2015
30
In how many ways can 4 letters be posted into 3 letters
boxes ?
(1)32 (2) 64 (3) 27 (4)81
Each letter can be posted in 3 ways.
Four letters can be posted in 3 x 3 x 3 x 3 or 81 ways.
25

COMPOUND EVENT
12 June 2015
31
When two coins are tossed and a cubical dice is rolled,
then the total outcomes for the compound event is?
(1)42 (2) 24 (3) 28 (4)10
When two coins are tossed there are 4 possible
outcomes.
When a dice is rolled there are 6 possible outcomes.
Hence there are 4 x 6 = 24 outcomes.
28

Permutation level two

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