IIT JEE

The number of zeroes at the end of (127)! is

(A) 31

(B) 30

(C) 0

(D) 10

Number of zeroes =

= 25 + 5 + 1 = 31

The value of the expression ^{k-1}C_{k-1}
+^{ k}C_{k-1} + . . . + ^{n+k-2}C_{k-1} is .

(A) ^{n+ k-1}C_{k+1}

(B) ^{n+ k-1}C_{k-1}

(C) ^{n+ k}C_{k}

(D) none of these

We have ^{k-1}C_{k-1}
+^{ k}C_{k-1} + . . . + ^{n+k-2}C_{k-1}.

= ^{k}C_{k} +^{ k}C_{k-1}
+^{ k+1}C_{k-1} . . . + ^{n+k-2}C_{k-1} [
since ^{k}C_{k} = 1 = ^{k-1}C_{k-1}].

= ^{k+1}C_{k} +^{
k+1}C_{k-1} +. . . . +^{ n+k-2}C_{k-1} [ since ^{k}C_{k}
+^{ k}C_{k-1}= ^{k+1}C_{k}] .

= ^{k+2}C_{k} + . . .
. + ^{n+k-2}C_{k-1} = . . .

= ^{n+k-2}C_{k} +^{
n+k-2}C_{k-1} =^{n+k-1}C_{k}.

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

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 ^{3}C_{1}
ways, there are ^{3}C_{1} (9´9)
= 3´ 9^{2} such
numbers. Next, 3 can occur exactly two places in (^{3}C_{2})
(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 ´ 9^{2}) + 2´(3´9)
+ 3´1 = 300.

The number of natural numbers which
are less than 2.10^{8} and which can be written by means of the
digits 1 and 2 is .

(A) 772

(B) 870

(C) 900

(D) 766

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 2^{2} two-digit such numbers and so
on.

There are 2^{8}
eight-digit such numbers. All the digit numbers beginning with 1 and written by
means of 1 and 2 are smaller than 2.10^{8}. Thus, there are 2^{8}
such nine-digit numbers. .

Hence the required number of numbers is

2 + 2^{2} +
2^{3}+ …. + 2^{8} + 2^{8} = + 2^{8} = 2^{9
}– 2 + 2^{8} = 766.

In a certain test,
there are n questions. In this test 2^{n-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

The number of
students answering exactly i ( 1£
i £ n-1) questions
wrongly is 2^{n-i} – 2^{n-i-1}. The number of students
answering all n questions wrongly is 2^{0} . Thus, the total number
of wrong answers is .

1(2^{n-1} –
2^{n-2}) +2( 2^{n-2} – 2^{n-3}) + . . . + ( n-1)( 2^{1}
- 2^{0}) +n(2^{0}).

= 2^{n-1}
+ 2^{n-2} + 2^{n-3} + . . . + 2^{0} = 2^{n} –
1.

Thus 2^{n}
–1 = 2047

&⇒ 2^{n} =
2048 = 2^{11}

&⇒ n = 11 .

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

One possible arrangement is

Three
such arrangements are possible. Therefore, the number of ways is (^{5}C_{2})
(^{3}C_{2}) (^{1}C_{1}) (3) = 90 .

The other possible arrangements

Three
such arrangements are possible. In this case, the number of ways is ( ^{5}C_{1})
( ^{4}C_{1}) ( ^{3}C_{3}) (3) = 60 .

Hence, the total number of ways is 90 + 60 = 150.

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

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 = ^{10}P_{2}
+1 (including 1).

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

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

Let A be the set
of 4-digit numbers a_{1}a_{2}a_{3}a_{4} where
a_{1}> a_{2}> a_{3}> a_{4}, then
n(A) is equal to

(A) 126

(B) 84

(C) 210

(D) none of these

Any selection of four digits from the
ten digits 0, 1, 2, 3, . . . , 9 gives one such number. So, the required
number of numbers = ^{10}C_{4} = 210 .

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) k^{k}

(C) 2^{k} –
1

(D) 2^{k}

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) = k^{k} . .