1

### JEE Main 2016 (Online) 10th April Morning Slot

The sum $\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)}$ is equal to :
A
(11)!
B
10 $\times$ (11!)
C
101 $\times$ (10!)
D
11 $\times$ (11!)

## Explanation

$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right)} \times r!$

$= \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2} - 2r} \right]\,.\,r!}$

$= \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2}\,.\,r!\,\, - \,\,2r\,.\,r!} \right]}$

$= \sum\limits_{r = 1}^{10} {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,\,2\sum\limits_{r = 1}^{10} {r\,.\,r!} }$

$= \sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,r\,\,.\,\,r!} \right]} - \sum\limits_{r = 1}^{10} {r\,\,.\,\,r!}$

$= \left[ {\left( {2.2! - 1.1!} \right) + \left( {3.3! - 2.2!} \right) + .... + \left( {11.11! - 10.10!} \right)} \right]$

$- \sum\limits_{r = 1}^{10} {r\,.\,r!}$

$=$   11.11! $-$ 1.1! $-$ $\sum\limits_{r = 1}^{10} {r\,.\,r!}$

$=$ (11.11! $-$ 1) $-$ $\sum\limits_{r = 1}^{10} {\left( {r + 1 - 1} \right)} \,.\,r!$

$=$ (11.11! $-$ 1) $-$ $\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)r! - r!} \right]}$

$=$ (11.11! $-$ 1) $-$ $\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)! - r!} \right]}$

$=$ (11.11! $-$ 1) $-$ [(2! $-$ 1!) + (3! $-$ 2!) + . . . .+ (11! $-$ 10!)]

$=$ (11.11! $-$ 1) $-$ (11! $-$ 1)

$=$ 11.11! $-$ 11!

$=$ 11! (11 $-$ 1)

$=$ $10\,.\,\left( {11!} \right)$
2

### JEE Main 2017 (Offline)

A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is:
A
468
B
469
C
484
D
485

## Explanation

X(7 Friends) Y(7 Friends)
Case 1 3 0 0 3
Case 2 0 3 3 0
Case 3 2 1 1 2
Case 4 1 2 2 1

In Case 1, Case 2, Case 3 and Case 4, total 6 friends are present and 3 from X and 3 from Y and among those 6 friend 3 are ladies and 3 are men in every case.

$\therefore$ No of ways 6 friends can be invited =

$({}^4{C_3} \times {}^3{C_0} \times {}^3{C_0} \times {}^4{C_3})$ + $({}^4{C_0} \times {}^3{C_3} \times {}^3{C_3} \times {}^4{C_0})$ + $\left( {{}^4{C_2} \times {}^3{C_1} \times {}^3{C_1} \times {}^4{C_2}} \right)$ + $\left( {{}^4{C_1} \times {}^3{C_2} \times {}^3{C_2} \times {}^4{C_1}} \right)$

= 16 + 1 + 324 + 144 = 485
3

### JEE Main 2017 (Online) 8th April Morning Slot

If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is :
A
44th
B
45th
C
46th
D
47th
4

### JEE Main 2017 (Online) 9th April Morning Slot

The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boy B1 and a particular girl G1 never sit adjacent to each other, is :
A
5 $\times$ 6!
B
6 $\times$ 6!
C
7!
D
5 $\times$ 7!

## Explanation

Number of ways = Total - when B1 and G1 sit together

Total ways to seat 8 people on round table = (8 - 1)! = 7!

When B1 and G1 sit together then assume B1 and G1 are one people, so total 7 people are there and among B1 and G1 they can sit 2! ways.

So total no of ways when B1 and G1 sit together
= (7 - 1)! $\times$ 2! = 6! $\times$ 2!

Number of ways = 7! - 6! $\times$ 2! = 6!$\times$(7 - 2) = 5 $\times$ 6!