Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

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 arrangement is :

A

at least 500 but less than 750

B

at least 750 but less than 1000

C

at least 1000

D

less than 500

From 6 different novels 4 novels can be chosen = $${}^6{C_4}$$ ways

And from 4 different dictionaries 1 can be chosen = $${}^3{C_1}$$ ways

$$\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C_4} \times {}^3{C_1}$$ ways

Let 4 novels are N_{1}, N_{2}, N_{3}, N_{4} and 1 dictionary is D_{1}.

Dictionary should be in the middle. So the arrangement will be like this

_ _ D_{1} _ _

On those 4 blank places 4 novels N_{1}, N_{2}, N_{3}, N_{4} can be placed. And 4 novels can be arrange $$4!$$ ways.

$$\therefore$$ Total no of ways = $${}^6{C_4} \times {}^3{C_1}$$$$ \times 4!$$ = 1080

And from 4 different dictionaries 1 can be chosen = $${}^3{C_1}$$ ways

$$\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C_4} \times {}^3{C_1}$$ ways

Let 4 novels are N

Dictionary should be in the middle. So the arrangement will be like this

_ _ D

On those 4 blank places 4 novels N

$$\therefore$$ Total no of ways = $${}^6{C_4} \times {}^3{C_1}$$$$ \times 4!$$ = 1080

2

MCQ (Single Correct Answer)

How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?

A

$$8.{}^6{C_4}.{}^7{C_4}$$

B

$$6.7.{}^8{C_4}$$

C

$$6.8.{}^7{C_4}$$.

D

$$7.{}^6{C_4}.{}^8{C_4}$$

This problem is solved using gap method. As here no 'S' is adjacent to each other so we have to put them in the gap. So first write all the letters other than 'S' such a way that there is a gap between two letters.

Given word is MISSISSIPPI.

Here, I = 4 times, S = 4 times, P = 2 times, M = 1 time

_M_I_I_I_I_P_P_

Those seven letters M, I, I, I, I, P, P can be arranged in $${{7!} \over {4!2!}}$$ ways

Those seven letters creates 8 gaps and we have to choose 4 gaps from those 8 gaps to put those four 'S' letters.

This can be done $${}^8{C_4}$$ ways.

After placing those four 'S' letters we can arrange them in $${{4!} \over {4!}}$$ ways.

Therefore, required number of words

$$ = {{7!} \over {4!2!}} \times {}^8{C_4} \times {{4!} \over {4!}}$$

$$ = {{7\,.\,6!} \over {4!4!}} \times {}^8{C_4}$$

$$ = 7\,.\,{}^6{C_4}\,.\,{}^8{C_4}$$

3

MCQ (Single Correct Answer)

In a shop there are five types of ice-cream available. A child buys six ice-cream.

** Statement - 1: ** The number of different ways the child can buy the six ice-cream is $${}^{10}{C_5}$$.

** Statement - 2: ** The number of different ways the child can buy the six ice-cream is equal to the number of different ways of arranging 6 A and 4 B's in a row.

A

Statement - 1 is false, Statement - 2 is true

B

Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1

C

Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1

D

Statement - 1 is true, Statement - 2 is false

Here n = 6 ice-cream

p = 5 types of ice-cream

Each ice-cream belongs to one of the 5 ice-cream type. So chosen 6 ice-crean can be divide into 5 types of ice-cream.

$$ \therefore $$ The number of different ways the child can buy the six ice-cream is = $${}^{6 + 5 - 1}{C_{5 - 1}}$$ = $${}^{10}{C_4}$$

$$ \therefore $$

Number of different ways of arranging 6 A and 4 B's in a row

= $${{10!} \over {6!4!}} = {}^{10}{C_4}$$

$$ \therefore $$

4

MCQ (Single Correct Answer)

The set S = {1, 2, 3, ........., 12} is to be partitioned into three sets A, B, C of equal size. Thus $$A \cup B \cup C = S,\,A \cap B = B \cap C = A \cap C = \phi $$. The number of ways to partition S is

A

$${{12!} \over {{{(4!)}^3}}}\,\,$$

B

$${{12!} \over {{{(4!)}^4}}}\,\,$$

C

$${{12!} \over {3!\,\,{{(4!)}^3}}}$$

D

$${{12!} \over {3!\,\,{{(4!)}^4}}}$$

The total number of ways is

$${}^{12}{C_4} \times {}^{12 - 4}{C_4} \times {}^{12 - 4 - 4}{C_4} = {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = {{12!} \over {{{(4!)}^3}}}$$

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Complex Numbers

Quadratic Equation and Inequalities

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Matrices and Determinants

Vector Algebra and 3D Geometry

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Properties of Triangle

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Straight Lines and Pair of Straight Lines

Circle

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Application of Derivatives

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