1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Evening Slot

The number of four letter words that can be formed using the letters of the word BARRACK is :
A
120
B
144
C
264
D
270

Explanation

Case 1 :

When all the four letters different then no of words
= 5C4 $$ \times $$4!

Case 2 :

When out of four letters two letters are R and other two different letters are chosen from B, A, C, K then the no of words
= 4C2 $$ \times $$ $${{4!} \over {2!}}$$ = 72

Case 3 :

When out of four letters two letters are A and other two different letters are chosen from B, R, C, K then the no of words
= 4C2 $$ \times $$ $${{4!} \over {2!}}$$ = 72

Case 4 :

When word is formed using two R and two A then number of words
= $${{4!} \over {2!2!}}$$ = 6

So, total number of 4 letters words possible = 120 + 72 + 72 + 6 = 270
2
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is :
A
24
B
30
C
36
D
48

Explanation

Here number should be divisible by 3, that means sum of numbers should be divisible by 3.

Possible 4 digits among 0, 1, 2, 3, 4 which are divisible by 3 are

(1)$$\,\,\,\,$$ (0, 2, 3, 4) Sum of digits = 0 + 2 + 3 +4 = 9 (divisible by 3)

(2) $$\,\,\,\,$$ (0, 1, 2, 3) Sum of digits = 0 + 1 + 2 + 3 = 6 (divisible by 3)

Case 1 :

When 4 digits are (0, 2, 3, 4) then



$$\therefore\,\,\,\,$$ Total possible numbers = $$^3{C_1}$$ $$ \times $$ $$^3{C_1}$$ $$ \times $$ $$^2{C_1}$$ $$ \times $$ $$^1{C_1}$$

=    3 $$ \times $$ 3 $$ \times $$ 2 $$ \times $$ 1 = 18

Case 2 :

When 4 digits are (0, 1, 2, 3) then,



$$\therefore\,\,\,\,$$ Total possible number in this case = $$^2{C_1}$$ $$ \times $$ $$^3{C_1}$$ $$ \times $$ $$^2{C_1}$$ $$ \times $$ $$^1{C_1}$$

=    2 $$ \times $$ 3 $$ \times $$ 2 $$ \times $$ 1 = 12

$$\therefore\,\,\,\,$$ Total possible numbers will be = 18 + 12 = 30
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is :
A
500
B
350
C
200
D
300

Explanation

From 5 girls 2 girls can be selected

= 5C2 ways

From 7 boys 3 boys can be selected

= 7C3 way

$$ \therefore $$  Total number of ways we can select 2 girls and 3 boys

= 5C2 $$ \times $$ 7C3 ways

When two boys A and B are chosen in a team then one more boy will be chosen from remaining 5 boys.

So, no of ways 3 boys can be chosen when A and B should must be chosen = 5C1 ways

$$ \therefore $$  Total number of ways a team of 2 girl and 3 boys can be made where boy A and B must be in the team = 5C1 $$ \times $$ 5C2 ways

$$ \therefore $$  Required number of ways
= Total number of ways $$-$$ when A and B are always included.

= 5C2 $$ \times $$ 7C3 $$-$$ 5C1 $$ \times $$ 5C2

= 300
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is :
A
9
B
18
C
36
D
32

Explanation



Area = $${1 \over 2}$$ h. k = 50

h. k = 100

h. k = 22 . 52

Total divisors

= (2 + 1) (2 + 1) = 9

if   h > 0, k > 0

But   $${\matrix{ {h > 0,} & {k < 0} \cr {h < 0,} & {k > 0} \cr {h < 0,} & {k < 0} \cr } }$$

all are possible so that total no. of positive case

9 + 9 + 9 + 9 = 36

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