1
MCQ (Single Correct Answer)

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
2
MCQ (Single Correct Answer)

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)

$$=$$ $$10\,.\,\left( {11!} \right)$$
3
MCQ (Single Correct Answer)

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)
4 Ladies 3 Men 3 Ladies 4 Men
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
4
MCQ (Single Correct Answer)

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

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