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Q:

A Group consists of 4 couples in which each of the  4 persons have one wife each. In how many ways could they be arranged in a straight line such that the men and women occupy alternate positions?

A) 1152 B) 1278
C) 1296 D) None of these

Answer:   A) 1152

Explanation:

Case I :  MW  MW  MW  MW

Case II:  WM  WM  WM  WM

Let us arrange 4 men in 4! ways, then we arrange 4 women in \inline _^{4}\textrm{p}_{4} ways at 4 places either left of the men or right of the men. Hence required number of arrangements

                   \inline =4!\times ^{4}\textrm{p}_{4}+ 4!\times ^{4}\textrm{p}_{4}=2\times 576=1152

Q:

Out of seven consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ?

A) 25200 B) 25000
C) 25225 D) 24752
 
Answer & Explanation Answer: A) 25200

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

=(   ) = ( ) = 210.

Number of groups, each having 3 consonants and 2 vowels =210

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = (5  4 3  2 1)=120.

Required number of words = (210 120) = 25200.

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Q:

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated ?

A) 15 B) 20
C) 5 D) 10
 
Answer & Explanation Answer: B) 20

Explanation:

Since each number to be divisible by 5, we must have 5 0r 0 at the units place. But in given digits we have only 5.
So, there is one way of doing it.
Tens place can be filled by any of the remaining 5 numbers.So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1  5  4) = 20.

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Q:

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there ?

A) 205 B) 194
C) 209 D) 159
 
Answer & Explanation Answer: C) 209

Explanation:

We may have (1 boy and 3 girls)or(2boys and 2 girls)or(3 boys and 1 girl)or(4 boys).
Required number of ways = () + () + () + ()

= () +  +  + 

= (24+90+80+15)

= 209.

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Q:

In a bag, there are 8 red, 7 blue and 6 green flowers. One of the flower is picked up randomly. What is the probability that it is neither red nor green ?

   

A) Option A B) Option B
C) Option C D) Option D
 
Answer & Explanation Answer: A) Option A

Explanation:

Total number of flowers = (8+7+6) = 21.
Let E = event that the flower drawn is neither red nor green.
= event taht the flower drawn is blue.
 n(E)= 7
 P(E)=  

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Q:

A box contains 5 green, 4 yellow and 5 white pearls. Four pearls are drawn at random. What is the probability that they are not of the same colour ?

A)  B)  C)  D) 

A) Option A B) Option B
C) Option C D) Option D
 
Answer & Explanation Answer: D) Option D

Explanation:

Let S be the sample space. Then,
n(s) = number of ways of drawing 4 pearls out of 14
=  ways =  = 1001
Let E be the event of drawing 4 pearls of the same colour.
Then, E = event of drawing (4 pearls out of 5) or (4 pearls out of 4) or (4 pearls out of 5)

 +  +  = 5+1+5 =11

 P(E) = 

  Required probability =  

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