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(1) At least 750 but less than 1000

(2) At least 1000

(3) Less than 500

(4) At least 500 but less than 750

**[AIEEE 2009]**

**Sol.**(2)

The no. of ways to select 4 novels $\& 1$ dictionary from 6 different novels $\& 3$ different dictionary are $6 \mathrm{C}_{4} \times^{3} \mathrm{C}_{1}$

and to arrange these things in shelf so that dictionary is always in middle $-\mathrm{D}–\mathrm{are} 4 !$

Required No. of ways $^{6} \mathrm{C}_{4} \times^{3} \mathrm{C}_{1} \times 4 !=1080$

(1) 3 (2) 36 (3) 66 (4) 108

**[AIEEE-2010]**

**Sol.**(4)

Urn $\mathrm{A} \rightarrow 3$ Red balls

Urn $\mathrm{B} \rightarrow 9$ Blue balls

So the number of ways $=$ selection of 2 balls from urn $\mathrm{A} \& \mathrm{B}$ each.

$=^{3} \mathrm{C}_{2} \cdot^{9} \mathrm{C}_{2}=108$

**Statement – 1 :**The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is $^{9} \mathrm{C}_{3}$

**Statement – 2 :** The number of ways of choosing any 3 places from 9 different places is $^{9} \mathrm{C}_{3}$

(1) Statement-1 is true, Statement-2 is false.

(2) Statement-1 is false, Statement-2 is true

(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement- 1

(4) Statement-1 is true, Statement-2 is true; Statement-2 is **not** a correct explanation for Statement- 1

**[AIEEE-2011]**

**Sol.**(3)

$\mathrm{B}_{1}+\mathrm{B}_{2}+\mathrm{B}_{3}+\mathrm{B}_{4}=10$

$\mathrm{St}-1: \mathrm{B}_{1} \geq 1, \mathrm{B}_{2} \geq 1, \mathrm{B}_{3} \geq 1, \mathrm{B}_{4} \geq 1$

so no. of negative integers solution of equation

$\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4}=10-4=6$

$6+4-1 \mathrm{C}_{4-1}=9 \mathrm{C}_{3}$

$\mathrm{St}-2:$ selction of 3 places from out of

9 places $=^{9} \mathrm{C}_{3}$

Both statements are true and correct explaination

(1) $\mathrm{N}>190$

(2) $\mathrm{N} \leq 100$

(3) $100<\mathrm{N} \leq 140$

(4) $140<\mathrm{N} \leq 190$

**[AIEEE-2011]**

**Sol.**(2)

$\begin{aligned} \mathrm{N}=10 \mathrm{C}_{3}-^{6} \mathrm{C}_{3} &=\frac{10 \times 9 \times 8}{3 \times 2 \times 1}-\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \\ &=120-20=100 \end{aligned}$

$\mathrm{N} \leq 100$

(1) 879 (2) 880 (3) 629 (4) 630

**[AIEEE-2012]**

**Sol.**(1)

$\mathrm{W}^{10}, \mathrm{G}^{9}, \mathrm{B}^{7}$

selection of one or more balls

$=(10+1)(9+1)(7+1)-1$

$=11 \times 10 \times 8-1=879$

(1) 256 (2) 220 (3) 219 (4) 211

**[JEE (Main)-2013]**

**Sol.**(3)

$\mathrm{n}(\mathrm{A})=2, \mathrm{n}(\mathrm{B})=4, \mathrm{n}(\mathrm{A} \times \mathrm{B})=\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})=8$

Now subsets having 3 or more elements.

$\mathrm{S} \mathrm{C}_{3}+^{8} \mathrm{C}_{4}+\ldots .+^{8} \mathrm{C}_{8}=2^{8}-^{8} \mathrm{C}_{0}-^{8} \mathrm{C}_{1}-^{8} \mathrm{C}_{2}=256-37=219$

(1) 7 (2) 5 (3) 10 (4) 8

**[JEE (Main)-2013]**

**Sol.**(2)

$\mathrm{T}_{\mathrm{n}}=\mathrm{n} \mathrm{C}_{3} \Rightarrow \mathrm{n}+1 \mathrm{C}_{3}-^{\mathrm{n}} \mathrm{C}_{3}=10$

$(\mathrm{n}+1) \mathrm{n}(\mathrm{n}-1)-\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)=60$

$\mathrm{n}(\mathrm{n}-1)=20$

$\mathrm{n}=5$

(1) 820 (2) 780 (3) 901 (4) 861

**[JEE (Main)-2015]**

**Sol.**(2)

(1) 275 (2) 510 (3) 219 (4) 256

**[JEE (Main)-2015]**

**Sol.**(3)

$\mathrm{n}(\mathrm{A})=4, \mathrm{n}(\mathrm{B})=2$

$\mathrm{n}(\mathrm{A} \times \mathrm{B})=8$

Number of subsets having atleast 3 elements

$=2^{8}-\left(1+^{8} \mathrm{C}_{1}+^{8} \mathrm{C}_{2}\right)=219$

(1) 120 (2) 72 (3) 216 (4) 192

**[JEE (Main)-2015]**

**Sol.**(4)

Number of 4 digit numbers greater than 6000 is $\frac{3}{2} \times 4 \times 3 \times 2=72$

Number of 5 digit numbers greater than 6000 is $5 !=120$

So total number of numbers $=72+120=192$

**[JEE (Main)-2016]**

**Sol.**(1)

Total number of words which can be formed using all the letters of the word ‘SMALL’

$=\frac{5 !}{2 !}=60$

Now, $60^{\text {th }}$ word is $\rightarrow$ SMLLA

59th word is $\rightarrow$ SMALL

58th word is $\rightarrow$ SMALL

(1) 484 (2) 485 (3) 468 (4) 469

**[JEE (Main)-2017]**

**Sol.**(2)

(1) less than 500

(2) at least 500 but less than 750

(3) at least 750 but less than 1000

(4) at least 1000

**[JEE (Main)-2018]**

**Sol.**(4)

$\begin{aligned} \text { Number of ways } &=\left(\begin{array}{c}{6} \\ {4}\end{array}\right)\left(\begin{array}{l}{3} \\ {1}\end{array}\right) 4 ! \\ &=15 \times 3 \times 24 \\ &=1080 \end{aligned}$