4
Q:

In the below word how many words are there in which R and W are at the end positions?

RAINBOW

A) 120 B) 180
C) 210 D) 240

Answer:   D) 240



Explanation:

When R and W are the first and last letters of all the words then we can arrange them in 5!ways. Similarly When W and R are the first and last letters of the words then the remaining letters can be arrange in 5! ways.

Thus the total number of permutations = 2 x 5!  = 2 x 120 = 240

Q:

When Dr. Ram arrives in his dispensary, he finds 12 patients waiting to see him. If he can see only one patients at a time,find the number of ways, he can schedule his patients if 3 leave in disgust before Dr. Ram gets around to seeing them.

A) 479001600 B) 79833600
C) 34879012 D) 67800983
 
Answer & Explanation Answer: B) 79833600

Explanation:

There are 12 - 3 = 9 patients.They can be seen in 12P9 = 79833600 ways

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3 6026
Q:

When John arrives in New York,he has eight stops to see, but he has time only to visit six of them.In how many different ways can he arrange his schedule in New York?

A) 20610 B) 24000
C) 20160 D) 21000
 
Answer & Explanation Answer: C) 20160

Explanation:

He can arrange his schedule in 8P6 = 20160 ways

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1 5026
Q:

Compute the sum of 4 digit numbers which can be formed with the four digits 1,3,5,7, if each digit is used only once in each arrangement.

A) 105555 B) 106665
C) 106656 D) 108333
 
Answer & Explanation Answer: C) 106656

Explanation:

The number of arrangements of 4 different digits taken 4 at a time is given by 4P4 = 4! = 24.All the four digits will occur equal number of times at each of the position,namely ones,tens,hundreds,thousands.

 

 

 

Thus,each digit will occur 24/4 = 6 times in each of the position.The sum of digits in one's position will be 6 x (1+3+5+7) = 96.Similar is the case in ten's,hundred's and thousand's places.

 

 

 

Therefore,the sum will be 96 + 96 x 10 + 96 x 100 + 96 x 100 = 106656

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9 13529
Q:

In how many different ways  can 3 students be associated with 4 chartered accountants,assuming that each chartered accountant can take at most one student?

A) 12 B) 36
C) 24 D) 16
 
Answer & Explanation Answer: C) 24

Explanation:

Number of permutations = 4P3 = 24

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1 5898
Q:

First,second and third prizes are to be awarded at an engineering fair in which 13 exhibits have been entered.In how many different ways can the prizes be awarded?

A) 1736 B) 1716
C) 1216 D) 1346
 
Answer & Explanation Answer: B) 1716

Explanation:

13P3= 13!/10! = 1716

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1 5762
Q:

In how many different ways can five persons stand in a line for a group photograph?

A) 120 B) 240
C) 360 D) 720
 
Answer & Explanation Answer: A) 120

Explanation:

This is the number of permutations of five things taken all at a time.

 

Therefore, answer = 5P5 = 120 ways

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2 6556
Q:

How many 3 letters words can be formed using the letters of the words hexagon?

A) 120 B) 210
C) 160 D) 200
 
Answer & Explanation Answer: B) 210

Explanation:

Since the word hexagon contains 7 different letters,the number of permutations is 7P3 = 7 x 6 x 5 =210

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4 10585
Q:

In how many ways can you choose one or more of 12 different candies?

A) 4054 B) 4050
C) 4095 D) 4059
 
Answer & Explanation Answer: C) 4095

Explanation:

Each candy can be dealt with in two ways.It can be chosen or not chosen.This will give 2 possibilites for the first candy, 2 for the second, and so on.by multiplying the cases together we get 212. Since the case of no candy being selected is not an option, we have to subtract 1 from our answer.

 

Therefore,there are 212- 1 = 4095 ways of selecting one or more candies

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0 7862