1

### 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

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

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

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

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

$\therefore$  $K = {2^{25}}$
2

### 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!)
3

### JEE Main 2019 (Online) 12th January Morning Slot

Consider three boxes, each containing, 10 balls labelled 1, 2, … , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni, the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is :
A
164
B
240
C
82
D
120

## Explanation

Number of ways = 10C3 = 120
4

### JEE Main 2019 (Online) 12th January Evening Slot

There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is :
A
12
B
9
C
7
D
11

## Explanation

Let m-men, 2-women

mC2 $\times$ 2 = mC1 2C1 . 2 + 84

m2 $-$ 5m $-$ 84 = 0 $\Rightarrow$ (m $-$ 12) (m + 7) = 0

m = 12

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