1

### 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
2

### 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
3

### JEE Main 2019 (Online) 10th January Evening Slot

If  $\sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_r}.{}^{50 - r}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)} ,\,\,$ then K is equal to -
A
224
B
225$-$ 1
C
225
D
(25)2

## Explanation

$\sum\limits_{r = 0}^{25} {^{50}} {C_r}.{}^{50 - r}{C_{25 - r}}$

$= \sum\limits_{r = 0}^{25} {{{50!} \over {r!\left( {50 - r} \right)!}}} \times {{\left( {50 - r} \right)!} \over {\left( {25} \right)!\left( {25 - r} \right)!}}$

$= \sum\limits_{r = 0}^{25} {{{50!} \over {25!25!}} \times {{25!} \over {\left( {25 - r} \right)!\left( {r!} \right)}}}$

$= {}^{50}{C_{25}}\sum\limits_{r = 0}^{25} {^{25}} {C_r} = \left( {{2^{25}}} \right){}^{50}{C_{25}}$

$\therefore$  $K = {2^{25}}$
4

### JEE Main 2019 (Online) 11th January Evening Slot

The number of functions f from {1, 2, 3, ...., 20} onto {1, 2, 3, ...., 20} such that f(k) is a multiple of 3, whenever k is a multiple of 4, is :
A
65 $\times$ (15)!
B
56 $\times$ 15
C
(15)! $\times$ 6!
D
5! $\times$ 6!

## Explanation

f(k) = 3m (3, 6, 9, 12, 15, 18)

for k = 4, 8, 12, 16, 20

6.5.4.3.2 ways

For rest numbers 15! ways

Total ways = 6! (15!)