Q. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangements is â€“
(1) At least 750 but less than 1000
(2) At least 1000
(3) Less than 500
(4) At least 500 but less than 750

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**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 $^{6} \mathrm{C}_{4} \times^{3} \mathrm{C}_{1} \times 4 !=1080$

Q. There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is –
(1) 3 (2) 36 (3) 66 (4) 108

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

Q.

**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]****Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...**

**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

Q. There are 10 points in a plane, out of these 6 are collinear. If N is the number of triangles formed by joining these points, then :
(1) $\mathrm{N}>190$
(2) $\mathrm{N} \leq 100$
(3) $100<\mathrm{N} \leq 140$
(4) $140<\mathrm{N} \leq 190$

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

Q. Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is â€“
(1) 879 (2) 880 (3) 629 (4) 630

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**Sol.**(1) $\mathrm{W}^{10}, \mathrm{G}^{9}, \mathrm{B}^{7}$ selection of one or more balls $=(10+1)(9+1)(7+1)-1$ $=11 \times 10 \times 8-1=879$

Q. Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A Ã— B having 3 or more elements is
(1) 256 (2) 220 (3) 219 (4) 211

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

Q. Let $\mathrm{T}_{\mathrm{n}}$ be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If $\mathrm{T}_{\mathrm{n}+1} \mathrm{T}_{\mathrm{n}}=10,$ then the value of $\mathrm{n}$ is :
(1) 7 (2) 5 (3) 10 (4) 8

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**Sol.**(2) $\mathrm{T}_{\mathrm{n}}=\mathrm{n} \mathrm{C}_{3} \Rightarrow \mathrm{n}+1 \mathrm{C}_{3}-^{\mathrm{n}} \mathrm{C}_{3}=10$ $(\mathrm{n}+1) \mathrm{n}(\mathrm{n}-1)-\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)=60$ $\mathrm{n}(\mathrm{n}-1)=20$ $\mathrm{n}=5$

Q. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0) is :
(1) 820 (2) 780 (3) 901 (4) 861

**[JEE (Main)-2015]**
Q. Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A Ã— B, each having at least three elements is :
(1) 275 (2) 510 (3) 219 (4) 256

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

Q. The number of integers greater than 6000 that can be formed, using the digits 3,5,6,7 and 8 without repetition, is :
(1) 120 (2) 72 (3) 216 (4) 192

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**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 $=72+120=192$

Q. If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is:

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**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

Q. 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 :
(1) 484 (2) 485 (3) 468 (4) 469

**[JEE (Main)-2017]**
Q. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is-
(1) less than 500
(2) at least 500 but less than 750
(3) at least 750 but less than 1000
(4) at least 1000

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

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