01.Five-digit
numbers are formed using only 0, 1, 2, 3, 4 exactly once. What is the
difference between the greatest and smallest numbers that can be
formed?
(a) 19800
(b) 41976
(c) 32976
(d) None of these
02.In how many ways can Eight Directors, Vice-Chairman and Chairman of a firm be seated at a round table, if the Chairman has to sit between the Vice-Chairman and a Director?
(a) 9! x 2
(b) 2 x 8!
(c) 2 x 7!
(d) None of these
03. A man has 9 friends: 4 boys and 5 girls. In how many ways can he invite them, if there have to be 3 exactly girls in the invitees?
(a) 320
(b) 160
(c) 80
(d) 200
04.
Boxes
numbered 1, 2, 3, 4 and 5 are kept in a row and they which are to be
filled with either a red or a blue ball, such that no two adjacent boxes
can be filled with blue balls. Then, how many different arrangements
are possible, given that all balls of a given colour are exactly
identical in all respects?
(a) 8
(b) 10
(c) 15
(d) 22
05.
A,
B, C, D are four towns, any three of which are non-collinear. Then, the
number of ways to construct three roads each joining a pair of towns so
that the roads do not form a triangle is?
(a) 7
(b) 8
(c) 9
(d) 24
06.
If
a 4-digit number is formed with digits 1, 2, 3 and 5. What is the
probability that the number is divisible by 25, if repetition of digits
is not allowed?
(a) 1/12
(b) 1/4
(c) 1/6
(d) None of these
07. A
five digit number is formed using digits 1, 3, 5, 7 and 9 without
repeating any one of them. What is the sum of all such possible numbers?
(a) 6666600
(b) 6666660
(c) 6666666
(d) None of these
08.139
persons have signed for an elimination tournament. All players are to
be paired up for the first round, but because 139 is an odd number one
player gets a bye, which promotes him to the second round, without
actually playing in the first round. The pairing continues on the next
round, with a bye to any player left over. If the schedule is planned so
that a minimum number of matches is required to determine the champion,
the number of matches which must be played is
(a) 136
(b) 137
(c) 138
(d) 139
09.A
box contains 6 red balls, 7 green balls and 5 blue balls. Each ball is
of different size. The probability that the red ball selected is the
smallest red ball?
(a) 1/18
(b) 1/3
(c) 1/6
(d) 2/3
10. A
group of 630 children is arranged in rows for a group photograph
session. Each row contains three fewer children than the row in front of
it. What number of rows is not possible?
(a) 3
(b) 4
(c) 5
(d) 6
Answers :
1. C
2. B
3. B
4. D
5. D
6. A
7. A
8. C
9. C
10. D
Permutation And Combinations 02
01. In
a chess competition involving some boys and girls of a school, every
student had to play exactly one game with the every other student. It
was found that in 45 games both the players were girls and in 190 games
both were boys. The number of games in which one player was a boy and
the other was a girl is?
(a) 200
(b) 216
(c) 235
(d) 256
02.
A
new flag is to be designed with six vertical stripes using some or all
of the colours yellow, green, blue and red. Then, the number of ways
this can be done such that no two adjacent stripes have the same colour
is?
(a) 12 x 81
(b) 16 x 192
(c) 20 x 125
(d) 24 x 216
03.
An
intelligence agency forms a code of two distinct digits selected from
0, 1, 2, …. 9 such that the first digit of the code is non-zero. The
code, handwritten on a slip, can however potentially create confusion,
when read upside down- for example, the code 91 may appear as 16. How
many codes are there for which no such confusion can arise?
(a) 80
(b) 78
(c) 71
(d) 69
04.
How many numbers can be made with digits 0, 7, 8 which are greater than 0 and less than a million?
(a) 496
(b) 486
(c) 1084
(d) 728
05.
In
how many ways is it possible to choose a white square and a black
square on a chess-board so that the squares must not lie in the same row
or column?
(a) 56
(b) 896
(c) 60
(d) 768
06.
How
many four-letter computer passwords can be formed using only the
symmetric letters. (no repetition allowed) (Symmetric letters :- A, H,
I, M, O, T, U, V, W, X AND Z)
(a) 7920
(b) 330
(c) 14640
(d) 419430
07.
How
many three-letter computer password can be formed with at least one
symmetric letter ? (Symmetric letters: A, H, I, M, O, T, U, V, W, X and
Z)
(a) 990
(b) 2730
(c) 12870
(d) 1560000
08.
For
a scholarship, at the most n candidates out of 2n+1 can be selected.
If the number of different ways of selection of at least one candidate
is 63, the maximum number of candidates that can be selected for the
scholarship is ?
(a) 3
(b) 4
(c) 6
(d) 5
09.
Ten
points are marked on a straight line and 11 points are marked on
another straight line. How many triangles can be constructed with
vertices from among the above points?
(a) 495
(b) 550
(c) 1045
(d) 2475
10.
How
many numbers can be formed from 1, 2, 3, 4, 5 (without repetition),
when the digit at unit’s place must be greater than the in the ten’s
place?
(a) 54
(b) 60
(c) 17
(d) 2 x 4!
Answers :Find Detailed Solutions at the end of the page
1. A
2. A
3. D
4. D
5. D
6. A
7. C
8. A
9. C
10. B
Set 1 Detailed Solutions
1.
Greatest five digit number : 43210
Smallest five digit number : 10234
Difference = 43210-10234= 32976
(c)
2.
We consider vice-chairman and the chairman as 1 Unit. Now, 9 persons
can be arranged along a circular table in 8! ways. And vice-chairman and
chairman can be arranged in 2 different ways. Hence required number of
ways=2 x 8!
(b)
3. 3 Girls can be selected out of 5 girls in 5C3 ways.
Since number of boys to be invited is not given, hence out of 4 boys, he can invite them (2)4 ways.
Hence required number of ways is = 5C3 x (2)4 = 160
(b)
4.Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls=25=32.
Two adjacent boxes with blue can be got in 4 ways.
i.e. (1, 2) (2, 3) (3, 4) (4, 5)
Three adjacent boxes with blue can be got in 3 ways
i.e. (1, 2, 3), (2, 3, 4) and (3, 4, 5)
Four adjacent boxes with blue can be be got in 2 ways.
i.e. (1234) , (2345)
and five boxes with blue can be got in 1 way.
Hence total number of ways of filling the boxes such that adjacent boxes have blue
(4 + 3 + 2 + 1)=10
(d)
5.
To construct 2 roads, three towns can be selected out of 4 in 4 x 3 x 2
= 24 ways. Now, if the third road goes from the third town to the first
town, a triangle is formed and if it goes to the fourth town, a
triangle is not formed. So there are 24 ways to form a triangle and 24
ways of avoiding a triangle.
(d)
6.
Total number of 4 digit numbers that can be formed = 4!. If the number
is divisible by 25, then the last two digit are 25. So the first two
digits can be arranged in 2! ways.
Hence required probability = 2!/4! = 1/12
(a)
7.
Keeping one digit in fixed position, other four can be arranged in 4!
ways= 24 ways. Thus each of the 5 digits will occur in each of the five
place 4! times. Hence the sum of digits in each position is 24(1 + 3 + 5
+ 7 + 9) = 600. So, the sum of all numbers
=600(1+10+100+1000+10000)=6666600
(a)
8. Required number of matches played will be (139 – 1)=138
(c)
9. Required probability is = 1/6
(c)
10. Let the number of students in the front row be x and the number of rows be n.
Hence, number of students in the next rows would be (x – 3) , (x – 6), (x – 9),……. and so on.
Now we have to check for each value of n=3, 4, 5, 6
Firstly take n=3
x + (x – 3) + (x – 6) =630
» 3x = 639 i.e. x = 213 ( Thus, n=3 is possible)
Likewise if n=4
x + (x – 3) + (x – 6)+(x – 9) =630
» 4x – 18 = 630 i.e. x = 162 ( Thus, n=4 is possible)
Likewise if n=5
x + (x – 3) + (x – 6)+(x – 9) +( x -12) =630
» 5x – 30 = 630 i.e. x = 132 ( Thus, n=5 is possible)
Likewise if n=6
x + (x – 3) + (x – 6)+(x – 9)+(x -12)+(x – 15)=630
» 6x – 45 = 630 i.e. x =112.5 (NOT AN INTEGER) ( Thus, n=6 is not possible)
(d)
Set 2 Detailed Solutions
1. Let there be m boys and n girls.
Then nC2 = 45 » n(n – 1)=90» n =10
mC2 = 190 » m(m – 1) = 380 » m=20
Number of games played between one boy and one girl
= 10C1 x 20C1= 10 x 20=200
(a)
2.
Any of the 4 colours can be chosen for the first stripe. Any of the
remaining 3 colours can be used for the second stripe. The stripe can
again be colored in 3 ways. ( We can repeat the colour of the first
stripe, but not use the colour of the second stripe).
Similarly, there are 3 ways to colour each of the remaining stripes.
» The number of ways the flag can be coloured is
4 x (3)5 = (12)(3)4
(a)
3.
The available digits are 0, 1, 2 …………… 9. The first digit can be chosen
in 9 ways ( 0 not acceptable ), the second digit can be accepted in 9
ways ( digits repetition not allowed). Thus the code can be made in 9 x
9=81 ways.
Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways.
Total number of ways confusion can arise = 4 x 3=12
Thus, required answer= 81-12=69
(d)
4.
Number of ways for selecting single digit=2
Number of ways for selecting two digit= 2 x 3=6
Number of ways for selecting three digit= 2 x 3 x 3=18
Number of ways for selecting four digit= 2 x 3 x 3 x 3=54
Number of ways for selecting five digit=2 x 3 x 3 x 3 x 3=162
Number of ways for selecting six digit= 2 x 3 x 3 x 3 x 3 x 3=486
Hence, total number of ways = (2 + 6 + 18 + 54 + 162 + 486)=728
(d)
5.
There are 32 black and 32 white square on a chess-board then number of
ways in choosing one white and one black square on the chess.
32C1 x 32C1= 32 x 32=1024
Number of ways in which square lies in the same row
White square=4
Black square=4
Number of rows=8
4C1 x 4C1 x 8 = 128
»Number of ways in which square lies in the same column= 128
Total number in which square lie on the same row or same column= 128 + 128=256.
(d)
6. Ist place of the four letter password can be filled in 11 ways.
IInd place of four letter password can be filled in 10 ways.
IIIrd place of four letter password can be filled in 9 ways.
IVth place of four letter password can be filled in 8 ways.
Hence, required number of ways= 11 x 10 x 9 x 8=7920 ways
(a)
7.
Three letter password from 26 letters can be selected in 26 x 25 x 24
ways. Three letter password from 15 asymmetric letters can be selected
in 15 x 14 x 13 ways.
Hence, three letter password with at least one symmetric letter can be made in (26 x 25 x 24)-(15 x 14 x 13)=12870 ways.
(c)
8. At lease one candidate out of (2n + 1) candidates can be selected in (2n+1 – 1) ways.
» 22n+1 – 1 = 63 » 22n+1 = 64 = (2)6» n=2.5
Since n cannot be a fraction. Hence n=3.
(a)
9. Required number of triangles formed
10C2 x 11 + 11C2 x 10= 45 x 11 + 55 x 10 = 1045
(c)
10.
The digit in the unit’s place should be greater than that in the ten’s
place. Hence, if digit 5 occupies the unit place then remaining four
digits need not to follow any order.
Hence required number of ways = 4!
However,
if digit 4 occupies the unit place then 5 cannot occupies the ten’s
positions. Hence, digits at the ten’s place will be one among 1, 2 or 3.
This can happen in 3 ways. The remaining 3 digits can be filled in the
remaining three places in 3! ways. Hence in all we have (3 x 3!) numbers
ending in 4.
Similarly,
if we have 3 in the unit’s place, the ten’s place can be either 1 or 2.
This can happen in 2 ways. The remaining 3 digits can be arranged in
the remaining 3 places in 3! ways. Hence we will have (2 x 3!) numbers
ending in 3. Similarly, we can find that there will be 3! numbers
ending in 2 and no number ending with 1. Hence total number of numbers
= 4! + (3) x 3! + (2)3! + 3!
= 4! + 6 x 3! = 24 + (6 x 6) = 60
(b)
