Q1. How many three digits numbers can be formed using digits 3,6,9?
(a). 22 (b). 27
(c). 20 (d). 25
Q2. How many odd numbers are there less than 10000 using digits 0,2,3,9
(a). 128 (b). 127
(c). 120 (d). 125
Q3. There are 8 people in the bus,3 get off on stop 1, 2 on stop 2, and rest on 3, how many different ways this can happen?
(a). 550 (b). 500
(c). 560 (d). 565
Q4. In how many ways can 7 Irish and 7 Welshmen sit down at a round table, no 2 Welshmen being in consecutive positions?
(a). 3628890 (b). 3628870
(c). 3628820 (d). 3628800
Q5. From 6 men and 5 ladies, in how many ways five people can be chosen with at least one man?
(a). 454 (b). 512
(c). 376 (d). 462
Q6.
In how many ways 11 cricketers can be chosen from 6 bowlers, 4 wicket
keepers and 11 batsmen to have 5 batsmen, 5 bowlers and 1 wicket keeper?
(a). 11008 (b). 11088
(c). 46200 (d). 44242
Q7. Find the sum of all the four digit numbers which can be formed with the digits 1,2,3,4.
(a). 74000 (b). 74066
(c). 78992 (d). 74088
Q8.
An Army officer has 10 soldiers in his platoon; he takes four at a time
to the forest trip, as often as he can without taking the same four
soldiers together more than once. How often will he go himself to the
forest?
(a). 156 (b). 210
(c). 256 (d). 186
Q9. In how many ways can the letters of the word “sincity” be arranged such that the two i’s always come together?
(a). 720 (b). 120
(c). 660 (d). 360
Q10. How many triangles can be formed by joining 12 points, 5 of which are collinear?
(a). 156 (b). 210
(c). 256 (d). 186
Q11.
How many signals can be made by raising 4 flags of different colours
one above the other when any number of them may be used?
(a). 64 (b). 72
(c). 92 (d). 84
Q12. Ravi is having a party for 10 people. In how many ways can he invite the guests?
(a). 1023 (b). 1024
(c). 720 (d). 1100
Q13.
In a chess competition involving some boys and girls of a school, every
student had to play exactly one game with 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
other was a girl is (CAT)
(a). 200 (b). 216
(c). 135 (d). 256
Q14.1
A string of three English letters is formed as per the following rules.
The first letter is any vowel. The second letter is m, n or p. If the
second letter is m then the third letter is any vowel which is
different from the first letter. If the second letter is n then the
third letter is e or u….If the second letter is p then the third letter
is same as the first letter. How many strings of letters can possibly
be formed using the above rules?(CAT)
(a). 40 (b). 45
(c). 30 (d). 35
Q15.
There are 12 towns grouped into four zones with three towns per zone.
It is intended to connect the towns with telephone lines such that
every two towns are connected with three direct lines if they belong to
the same zone and with only one direct line otherwise. How many direct
telephone lines are required?(CAT)
(a). 72 (b). 90
(c). 96 (d). 144
Q16.
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 nonzero. 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?(CAT)
(a). 80 (b). 78
(c). 71 (d). 69
Q17.
In how many ways is it possible to choose a white square and a black
square on a chess board so that the square must not lie in the same row
or column?(CAT)
(a). 56 (b). 896
(c). 60 (d). 768
Ans1. (b) Three digit, three places = 3x3x3 = 27
Ans2. (a) Single Digit = 2(3,9)
Two Digit = 6
Three Digit = 24
Four Digit = 96
Total = 128
Ans3. (c) 8!/3!2!3! = 560
Ans4. (d). There
are 14 people to sit; let first Irish be seated, first person can sit
anywhere, therefore 6! Ways, and then the Welshmen have 7! Ways to take
7 positions
Total ways = 7! X 6! = 3628800
Putting 1 Englishman in a fixed position, the remaining 6 can be arranged in 6!
Ans5. (d). 6 Men + 5 Ladies, 10 people, five to be selected, which is 11C5
The scenario in which no man is selected, 5 women, all 5 to be selected 5C5 = 1
Here 11C5 - 5C5 = 462 ways
Ans6. (b) One wicketkeeper has to be chosen out of 4 which is 4C1 = 4
Five batsman have to be chosen out of 11 which is 11C5
Five bowlers have to be choose out of 6 which is 6C5 = 6
Total ways = 4 x 11C5 x 6 = 11088
Ans7. (d) Numbers that can be formed from 1,2,3,4 = 4! = 24
The least number is 1234 and the greatest is 4321
Sum = Average x total = (1234+4321)/2 x 24 = 74088
Ans8.(b) The visits of the officer = number of groups of four, as he visits with each group
Which is four soldiers going out of 10 = 10C4 = 210 times
Ans9.(a) Since
two i’s have to come together, total words are 7, so there will be a
set of six now(2 i’s and 5 others), which is 6! Ways, now two i’s can
arrange themselves in only one way, therefore ways are 6! = 720
Ans10. (d) 3 points required for a triangle therefore 12C3, since 5 points are collinear, therefore these points will not form a triangle 5C3. Number of Triangles = 12C3 - 5C3 = 210
Ans11. (a) Flags can be used in any number, so from 1 to 4
Therefore no. signals = 4P1 + 4P2 + 4P3 + 4P4 = 64
Ans12 (c) 210 – 1 = 1023
Ans13. (a) For a match two people are required in case of boys = nC2 = 45 = n(n-1) / 1
n(n-1) = 90, n = 10 boys
In case of girls = nC2 = 190 = n(n-1) / 2 = 190
n (n-1) = 380, n = 20 girls
Matches between girls and boys are 10C1 x 20C1 = 10 x 20 = 200.
Ans14.1. (d) First place can be taken by 5 vowels
Case I – second place is 1(m) and third is 4(other vowels) = 5x1x4 = 20
Case II – second place is 1(m) and third is 2(2 vowels) = 5x1x2 = 10
Case III – second place is 1(m) and third is 1(same vowel) = 5x1x1 = 5
Total = 35
Ans14.2. (c) First
place can be taken by 5 vowels in only case II, In case I first place
will be taken by 4 and in Case III it will be taken by 1.
Case I – second place is 1(m) and third is 1(only e) = 4x1x1 = 4
Case II – second place is 1(m) and third is 1(only e) = 5x1x1 = 5
Case III – second place is 1(m) and third is 1(same vowel) = 1x1x1 = 1
Total = 10 ways
Ans15. (b) Total number of lines required for connections in each zone = 9 x 4 = 36
Each town connecting to town in different zone will require 3 x 3 = 9 lines
Selecting 2 out of 4 towns = 4C2 = 6
Lines required for connecting towns of different zones = 6 x 9 = 54
Total number of lines in all = 54 + 36 = 90.
Ans16. (c) Total codes which can be formed = 9 x 9 = 81 (distinct digits)
The digits which can confuse are 1,6,8,9. From these digits we can form the codes
= 4 x 3 = 12
Out of these 12 codes two numbers 69 and 96 will not create confusion. Therefore
( 12 – 2 ) = 10 codes will create confusion
Thus total codes without confusion = 81 – 10 = 71
Ans17.(a) In
a chess board , there are 8 rows and 8 columns. If you choose one row
or column you have to choose one less, as a box cannot be taken from
same row or column, Required number of ways = 8C1 x 7C1 = 8 x 7 = 56.
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