1

### JEE Main 2016 (Online) 9th April Morning Slot

If the four letter words (need not be meaningful ) are to be formed using the letters from the word “MEDITERRANEAN” such that the first letter is R and the fourth letter is E, then the total number of all such words is :
A
${{11!} \over {{{\left( {2!} \right)}^3}}}$
B
110
C
56
D
59

## Explanation

Here total no of different letters present are,

(1)   One M

(2)   Three E (E E E)

(3)   One D

(4)   One I

(5)   One T

(6)   Two R (R R)

(7)   Two A (A A)

(8)    Two N (N N)

In the four letter word first letter is R and last letter is E.

$\therefore$     Word is = R _ _ E

Now remaining letters are,

M, EE, D, I, T, R, AA, NN

Those 2 empty places can be filled with identical letters [EE, AA, NN] in 3 ways.

Or two empty places can be filled with distinct letters [M, E, D, I, T, R, A, N] in ${}^8{C_2} \times 2!$ ways.

$\therefore$   Total no of words = 3 + ${}^8{C_2} \times 2!$ = 59
2

### JEE Main 2016 (Online) 9th April Morning Slot

The value of $\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}}} \right)$ is equal to :
A
560
B
680
C
1240
D
1085

## Explanation

We know,

${{{}^n{C_r}} \over {{}^n{C_{r - 1}}}} = {{n - r + 1} \over r}$

$\therefore$    ${{^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}} = {{15 - r + 1} \over r} = {{16 - r} \over r}$

$\therefore$     $\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {^{15}{C_{r - 1}}}}} \right)$

$=$ $\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{16 - r} \over r}} \right)$

$= \sum\limits_{r = 1}^{15} {\left( {16r - {r^2}} \right)}$

$= 16\sum\limits_{r = 1}^{15} {r - \sum\limits_{r = 1}^{15} {{r^2}} }$

$= 16 \times {{15 \times 16} \over 2} - {{15 \times 16 \times 31} \over 6}$

$=$  120 $\times$ 16 $-$ 40 $\times$ 31

$=$ 680
3

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

If    ${{{}^{n + 2}C{}_6} \over {{}^{n - 2}{P_2}}}$ = 11, then n satisfies the equation :
A
n2 + 3n − 108 = 0
B
n2 + 5n − 84 = 0
C
n2 + 2n − 80 = 0
D
n2 + n − 110 = 0

## Explanation

${{{}^{n + 2}{C_6}} \over {{}^{n - 2}{P_2}}} = 11$

$\Rightarrow$   ${{\left( {n + 2} \right)!} \over {6!\,\left( {n - 4} \right)!}} = 11\,.\,{{\left( {n - 2} \right)!} \over {\left( {n - 4} \right)!}}$

$\Rightarrow$   (n + 2)!  =  11.6! (n $-$ 2)!

$\Rightarrow$   (n + 2) (n + 1) n (n $-$ 1)   =   11.6!

$\Rightarrow$   (n + 2) (n + 1) n (n $-$ 1) = 11 . 6 . 5 . 4 . 3 . 2 . 1

$\Rightarrow$   (n + 2) (n + 1) n (n $-$ 1)   =  11 . 10 . 9 . 8

$\therefore$   n = 9

This value of n satisfy the equation,

n2 + 3n $-$ 108 = 0
4

### 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)

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