Question.1
The number of zeroes at the end of (127)! is
(A) 31
(B) 30
(C) 0
(D) 10
Solution
Number of zeroes = 
= 25 + 5 + 1 = 31
Question.2
The value of the expression k-1Ck-1
+ kCk-1 + . . . + n+k-2Ck-1 is .
(A) n+ k-1Ck+1
(B) n+ k-1Ck-1
(C) n+ kCk
(D) none of these
Solution
We have k-1Ck-1
+ kCk-1 + . . . + n+k-2Ck-1.
= kCk + kCk-1
+ k+1Ck-1 . . . + n+k-2Ck-1 [
since kCk = 1 = k-1Ck-1].
= k+1Ck +
k+1Ck-1 +. . . . + n+k-2Ck-1 [ since kCk
+ kCk-1= k+1Ck] .
= k+2Ck + . . .
. + n+k-2Ck-1 = . . .
= n+k-2Ck +
n+k-2Ck-1 =n+k-1Ck.
Question.3
The number of times
of the digits 3 will be written when listing the integer from 1 to 1000 is
(A) 269
(B) 300
(B) 271
(D) 302
Solution
Since 3 does not
occur in 1000, we have to count the number of times 3 ocures when we
list the integers from 1 to 999. Any number between 1 and 999 is of the form
xyz where 0 £ x, y, z£ 9 . Let us first count the numbers
in which 3 occurs exactly once. Since 3 can occur at one place in 3C1
ways, there are 3C1 (9´9)
= 3´ 92 such
numbers. Next, 3 can occur exactly two places in (3C2)
(9) = 3 ´ 9 such numbers .
Lastly,
3 can occur in all three digits in one number only. Hence the number of times 3
occurs is 1 ´ (3 ´ 92) + 2´(3´9)
+ 3´1 = 300.
Question.4
The number of natural numbers which
are less than 2.108 and which can be written by means of the
digits 1 and 2 is .
(A) 772
(B) 870
(C) 900
(D) 766
Solution
The required numbers are 1, 2, 11,
12, 21, 22, . . . ,122222222. .
Let us calculate how
many numbers are these.
There are 2
one-digit such numbers. There are 22 two-digit such numbers and so
on.
There are 28
eight-digit such numbers. All the digit numbers beginning with 1 and written by
means of 1 and 2 are smaller than 2.108. Thus, there are 28
such nine-digit numbers. .
Hence the required
number of numbers is
2 + 22 +
23+ …. + 28 + 28 = 
+ 28 = 29
– 2 + 28 = 766.
Question.5
In a certain test,
there are n questions. In this test 2n-i students gave wrong answers
to at lest i question, where i = 1, 2, . . . , n .If the total number of
wrong answers given is 2047, then n is equal to.
(A) 10
(B) 11
(C) 12
(D) 13
Solution
The number of
students answering exactly i ( 1£
i £ n-1) questions
wrongly is 2n-i – 2n-i-1. The number of students
answering all n questions wrongly is 20 . Thus, the total number
of wrong answers is .
1(2n-1 –
2n-2) +2( 2n-2 – 2n-3) + . . . + ( n-1)( 21
- 20) +n(20).
= 2n-1
+ 2n-2 + 2n-3 + . . . + 20 = 2n –
1.
Thus 2n
–1 = 2047
&⇒ 2n =
2048 = 211
&⇒ n = 11 .
Question.6
Five balls of
different colours are to be placed in three boxes of different sizes. Each
box can hold all five balls. The number of ways in which we can place the
balls in the boxes ( order is not considered in the box) so that no box
remains empty is .
(A) 150
(B) 300
(C) 200
(D) none of these
Solution
One possible
arrangement is
Three
such arrangements are possible. Therefore, the number of ways is (5C2)
(3C2) (1C1) (3) = 90 .
The
other possible arrangements
Three
such arrangements are possible. In this case, the number of ways is ( 5C1)
( 4C1) ( 3C3) (3) = 60 .
Hence,
the total number of ways is 90 + 60 = 150.
Question.7
Let A = {x l x is a prime number and x <
30} .The number of different rational numbers whose numerator and
denominator belong to A is .
(A) 90
(B) 180
(C) 91
(D) none of these
Solution
A
= {2, 3, 5, 7, 11,13, 17, 19, 23, 29}. A rational number is made by taking
any two in any order . So, the required number of rational numbers = 10P2
+1 (including 1).
Question.8
The number of ways of
arranging six persons (having A, B, C and D among them) in a row so that A, B,
C and D are always in order ABCD (not necessarily together) is
(A) 4
(B) 10
(C) 30
(D) 720
Solution
The number of ways of
arranging ABCD is 4!. For each arrangement of ABCD, the number of ways of
arranging six persons is same. Hence required number is 
Question.9
Let A be the set
of 4-digit numbers a1a2a3a4 where
a1> a2> a3> a4, then
n(A) is equal to
(A) 126
(B) 84
(C) 210
(D) none of these
Solution
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 .
Question.10
Let S be the set of all functions
from the set A to the set A. If n(A) = k, then n(S) is .
(A) k!
(B) kk
(C) 2k –
1
(D) 2k
Solution
Each element of
the set A can be given the image in the set A in k ways. So, the
required number of functions, i.e. , n(S) = k´ k ´
. . .(k times) = kk . .
