1

### JEE Main 2021 (Online) 1st September Evening Shift

Let P1, P2, ......, P15 be 15 points on a circle. The number of distinct triangles formed by points Pi, Pj, Pk such that i +j + k $\ne$ 15, is :
A
12
B
419
C
443
D
455

## Explanation

Total number of triangles = ${}^{15}{C_3}$

i + j + k = 15 (Given) Number of possible triangles using the vertices Pi, Pj, Pk such that i + j + k $\ne$ 15 is equal to ${}^{15}{C_3}$ $-$ 12 = 443

Option (c)
2

### JEE Main 2021 (Online) 25th July Evening Shift

If ${}^n{P_r} = {}^n{P_{r + 1}}$ and ${}^n{C_r} = {}^n{C_{r - 1}}$, then the value of r is equal to :
A
1
B
4
C
2
D
3

## Explanation

${}^n{P_r} = {}^n{P_{r + 1}} \Rightarrow {{n!} \over {(n - r)!}} = {{n!} \over {(n - r - 1)!}}$

$\Rightarrow (n - r) = 1$ .....(1)

${}^n{C_r} = {}^n{C_{r - 1}}$

$\Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}$

$\Rightarrow {1 \over {r(n - r)!}} = {1 \over {(n - r + 1)(n - r)!}}$

$\Rightarrow n - r + 1 = r$

$\Rightarrow n + 1 = 2r$ ..... (2)

From (1) and (2), $2r - 1 - r = 1 \Rightarrow r = 2$
3

### JEE Main 2021 (Online) 18th March Morning Shift

The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is :
A
26664
B
122664
C
122234
D
22264

## Explanation

Total possible numbers using 1, 2, 2 and 3 is

= ${{4!} \over {2!}}$ = 12

When unit place is 1, the total possible numbers using remaining 2, 2 and 3 are

= ${{3!} \over {2!}}$ = 3

When unit place is 2, the total possible numbers using remaining 1, 2 and 3 are

= 3! = 6

When unit place is 3, the total possible numbers using remaining 1, 2 and 2 are

= ${{3!} \over {2!}}$ = 3

$\therefore$ Sum of unit places of all (3 + 6 + 3) 12 numbers is

= ( 1$\times$3 + 2$\times$6 + 3$\times$3)

Similarly,

When 10th place is 1, the total possible numbers using remaining 2, 2 and 3 are

= ${{3!} \over {2!}}$ = 3

When 10th place is 2, the total possible numbers using remaining 1, 2 and 3 are

= 3! = 6

When 10th place is 3, the total possible numbers using remaining 1, 2 and 2 are

= ${{3!} \over {2!}}$ = 3

$\therefore$ Sum of 10th places of all (3 + 6 + 3) 12 numbers is

= ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 10

Similarly,

Sum of 100th places of all (3 + 6 + 3) 12 numbers is

= ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 100

and Sum of 1000th places of all (3 + 6 + 3) 12 numbers is

= ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 1000

$\therefore$ Total sum = ( 1$\times$3 + 2$\times$6 + 3$\times$3) + ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 10

+ ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 100 + ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 1000

= (3 + 12 + 9) (1 + 10 + 100 + 1000) = 1111 $\times$ 24 = 26664
4

### JEE Main 2021 (Online) 17th March Evening Shift

If the sides AB, BC and CA of a triangle ABC have 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to :
A
240
B
360
C
333
D
364

## Explanation Total number of triangles

= ${}^{14}{C_3} - {}^3{C_3} - {}^5{C_3} - {}^6{C_3}$

= 364 – 31 = 333