1

### 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
2

### 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
3

### 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
4

### JEE Main 2019 (Online) 9th January Evening Slot

The number of natural numbers less than 7,000 which can be formed by using the digits 0, 1, 3, 7, 9 (repitition of digits allowed) is equal to :
A
374
B
372
C
375
D
250

## Explanation

Total no 1 digit numbers possible = 4 (allowed digits 1, 3, 7, 9)

Total no 2 digit numbers possible = 4$\times$5 = 20

Total no 3 digit numbers possible = 4$\times$5$\times$5 = 100

Total no 4 digit numbers possible = 2$\times$5$\times$5$\times$5 = 250

So the number of natural numbers less than 7,000 possible are
= 4 + 20 + 100 + 250 = 374

NEET