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:

Out of 6 green ball, 4 blue ball, in how many ways we selectone or more balls ?

A) 42 B) 34
C) 31 D) 22
 
Answer & Explanation Answer: B) 34

Explanation:

7 × 5 = 35

35 – 1 = 34

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6 3625
Q:

A box contains 5 green, 4 yellow and 3 white marbles. Threemarbles are drawn at random. What is the probability thatthey are not of the same colour ?

A) 40/44 B) 44/41
C) 41/44 D) 40/39
 
Answer & Explanation Answer: C) 41/44

Explanation:

n(E) = 5C3 + 4C3 + 3C3 = 10 + 4 + 1 = 15n(S) = 12C3 = 220

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Filed Under: Permutations and Combinations
Exam Prep: Bank Exams

9 1898
Q:

There are 5 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next two have 6 choices each?

A) 1112 B) 2304
C) 1224 D) 2426
 
Answer & Explanation Answer: B) 2304

Explanation:

Number of questions = 5
Possibilities of choices for each question 1 to 5 respectively = 4, 4, 4, 6, 6


Reuired total number of sequences

= 4 x 4 x 4 x 6 x 6

= 2304.

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20 5772
Q:

How many numbers of five digits can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits?

A) 120 B) 240
C) 256 D) 360
 
Answer & Explanation Answer: A) 120

Explanation:

Required number of 5 digit numbers can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits are 

5 x 4 x 3 x 2 x 1 = 120.

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8 2313
Q:

What is the value of 100P2 ?

A) 10000 B) 9900
C) 8900 D) 7900
 
Answer & Explanation Answer: B) 9900

Explanation:

Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged.

 

That can be done as,

 

100 P2  = 100!/(100 - 2)!

= 100 x 99 x 98!/98!

= 100 x 99 

= 9900.

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7 6737
Q:

In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels always come together?

A) 720 B) 1440
C) 1800 D) 3600
 
Answer & Explanation Answer: B) 1440

Explanation:

Given word is THERAPY.

Number of letters in the given word = 7

Number of vowels in the given word = 2 = A & E

Required number of different ways, the letters of the word THERAPY arranged such that vowels always come together is

6! x 2! = 720 x 2 = 1440.

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13 1932
Q:

In how many different ways the letters of the word 'TRANSFORMER' can be arranged such that 'N' and 'S' always come together?

A) 112420 B) 85120
C) 40320 D) 1209600
 
Answer & Explanation Answer: D) 1209600

Explanation:

Given word is TRANSFORMER.

Number of letters in the given word = 11 (3 R's)

 

Required, number of ways the letters of the word 'TRANSFORMER' can be arranged such that 'N' and 'S' always come together is

10! x 2!/3!

= 3628800 x 2/6

= 1209600

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7 2453
Q:

In how many ways the letters of the word 'CIRCUMSTANCES' can be arranged such that all vowels came at odd places and N always comes at end?

A) 1,51,200 ways. B) 5,04,020 ways
C) 72,000 ways D) None of the above
 
Answer & Explanation Answer: A) 1,51,200 ways.

Explanation:

In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E

Total = 13 letters

But last letter must be N

Hence, available places = 12

In that odd places = 1, 3, 5, 7, 9, 11

Owvels = 4

This can be done in 6P4 ways 

Remaining 7 letters can be arranged in 7!/3! x 2! ways

 

Hence, total number of ways = 6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.

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3 2809