Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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

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

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

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

2

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

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

$$ \therefore $$

3

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

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

4

MCQ (Single Correct Answer)

At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is

A

5040

B

6210

C

385

D

1110

A voter can give vote to either 1 candidate or 2 candidates or 3 candidates or 4 candidates.

Case 1 : When he give vote to only 1 candidate then no ways = $${}^{10}{C_1}$$

Case 2 : When he give vote to 2 candidates then no ways = $${}^{10}{C_2}$$

Case 3 : When he give vote to 3 candidates then no ways = $${}^{10}{C_3}$$

Case 4 : When he give vote to 4 candidates then no ways = $${}^{10}{C_4}$$

So, total no of ways he can give votes

= $${}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4}$$

= 385

**Note :** Here we use addition rule as he can vote any one of those four rules. Whenever there is "or" choices, we use addition rule.

Case 1 : When he give vote to only 1 candidate then no ways = $${}^{10}{C_1}$$

Case 2 : When he give vote to 2 candidates then no ways = $${}^{10}{C_2}$$

Case 3 : When he give vote to 3 candidates then no ways = $${}^{10}{C_3}$$

Case 4 : When he give vote to 4 candidates then no ways = $${}^{10}{C_4}$$

So, total no of ways he can give votes

= $${}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4}$$

= 385

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