IIT JEE

A teacher takes 3 children from her class to the zoo at a time as often as she can, but she does not take the same three children to the zoo more than once. She finds that she goes to the zoo 84 times more than a particular child goes to the zoo. The number of children in her class is .

(A) 12

(B) 10

(C) 60

(D) none of these .

The number of times the teacher goes to
the zoo = ^{n}C_{3}.

The number of times
a particular child goes to the zoo = ^{n-1}C_{2}.

From the question
, ^{n}C_{3} – ^{n-1}C_{2} = 84.

Or (n -1)(n –2)(n –3) = 6 ´ 84 = 9´ 8 ´ 7

&⇒ n – 1 = 9

&⇒ n = 10

Two teams are to play a series of 5 matches between them. A match ends in a win or loss or draw for a team. A number of people forecast the result of each match and no two people make the same forecast for the series of matches. The smallest group of people in which one person forecasts correctly for all the matches will contain n people, where n is .

(A) 81

(B) 243

(C) 486

(D) none of these

The smallest number of people = total number of possible forecasts

= total number of possible results

= 3´ 3 ´ 3´3 ´3 = 243.

In a plane there are two families of lines y = x +r, y = -x +r ,

where r ∈ { 0, 1, 2, 3, 4} . The number of squares of diagonals of length 2 formed by the lines is .

(A) 9

(B) 16

(C) 25

(D) none of these .

There are two sets of five parallel lines at equal
distances. Clearly, lines like l_{1}, l_{3}, m_{1}, m_{3}
form a squares whose diagonal’s length is 2.

So, the number of required squares = 3´ 3

{since choices are (l_{1}, l_{3}),

(l_{2}, l_{4}), (l_{3}, l_{5}) for one set,
etc)

In the next word cup of cricket there will be 12 teams, divided equally in two groups. Teams of each group will play a match againast each other. From each group 3 top teams will qualify for the next round. In this round each team will play against others once. Four top teams of this round will qualify for the semifinal round, where each team will play against the other three. Two top teams of this round will go to the final round, where they will play the best of three matches. The minimum number of matches in the next world cup will be .

(A) 54

(B) 53

(C) 38

(D) none of these

The number of matches in the first round
= ^{6}C_{2}+^{6}C_{2}. .

The number of matches in the next round = ^{6}C_{2}

The number of matches in the semifinal
round round = ^{4}C_{2}.

So, the required number of matches

= ^{6}C_{2}+^{6}C_{2}+^{6}C_{2}+^{4}C_{2}
+2 = 53

(Note: For “ best of three” at least two matches are played.).

There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum number of triangles with vertices at these points is .

(A) 3p^{2}(
p-1) +1

(B) 3p^{2}(
p-1)

(C) p^{2}(
4p –3)

(D) none of these

The number of triangles with vertices on different lines

= ^{p}C_{1}
´ ^{P}C_{1}
´^{p}C_{1} = p^{3}
.

The number of triangles with two vertices on one line and the third vertex on any one of the other two lines

= ^{3}C_{1
}{^{p}C_{2} ´^{2p}C_{1}} = 6p.

so, the required number of triangles

= p^{3} + 3p^{2}(p
-1) = p^{2}( 4p – 3)

The number of rational numbers , where p, q ∈ {1, 2, 3, 4, 5, 6,} is

(A) 23

(B) 32

(C) 36

(D) none of these

Distinct numbers of the form are 36 minus the number of numbers where greatest common divisor of p and q is more than 1. Such numbers are 23.

The number of ways in which 9 identical balls can be placed in three identical boxes is

(A) 55

(B)

(C)

(D) 12

If number of balls is a, b and c in different boxes,
then a + b + c = 9 Number of solutions is ^{11}C_{2} = 55.55
ways include those ways in this.

(i) a, b, c are same

(ii) two of a, b, c are equal

(iii) all a, b, c are distinct

Now a, b, c are same in exactly one way (a= b = c = 3). Also three are 12 ways out of 55 ways in which exactly two of a, b, c are same. These 12 will be counted as = 4 ways (as 3,3,1:3, 1, 3 and 1, 3, 3 are same ways. Thus total number of required ways is 1+4+7 = 12.

Number of ways in which 5 identical objects can be distributed in 8 persons such that no person gets more than one object is

(A) 8

(B) ^{8}C_{5}

(C) ^{8}P_{5}

(D) None of these

No. of ways = Coefficient of x^{5} in (x^{0}
+ x^{1})^{8} = ^{8}C_{5}.

The number of flags with three strips in order, that can be formed using 2 identical red, 2 identical blue and 2 identical white strips is

(A) 24

(B) 20

(C) 90

(D) 8

No.
required flags = 3! ´ coefficient of x^{3} in

= 6 ´4 = 24.

Let and be a variable vector such that are positive integers. If £ 12 then the number of values of is

(A) ^{12}C_{9}
–1

(B) ^{12}C_{3}

(C) ^{12}C_{9}

(D) none of these

If then from the question x, y, z are positive integers.

Also £ 12

&⇒
x +y + z £ 12. So, the
number of values of

= the number of positive integral solutions of ( x+y+z £ 12)

= ^{2}C_{2} +^{3}C_{2}
+ . . . + ^{11}C_{2}

= ^{3}C_{0} + ^{3}C_{1}
+^{4}C_{2}+ . . . + ^{11}C_{9} ( since ^{n}C_{r}
= ^{n}C_{n-r}).

= ^{4}C_{1} + ^{4}C_{2}
+ . . . +^{11}C_{9} = . . . = ^{12}C_{9} .