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Q:

What is the total no of ways of selecting atleast one item from each of the two sets containing 6 different items each?

A) 2856 B) 3969
C) 480 D) None of these

Answer:   B) 3969



Explanation:

We can select atleast one item from 6 different items = \inline (2^{6}-1)

Similarly we can select atleast one item from other set of 6 different items in \inline (2^{6}-1) ways.

\therefore Required number of ways = \inline (2^{6}-1)(2^{6}-1) 

                                         = \inline (2^{6}-1)^2 = 3969

Q:

To fill 8 vacancies there are 15 candidates of which 5 are from ST. If 3 of the vacancies are reserved for ST candidates while the rest are open to all, Find the number of ways in which the selection can be done ?

A) 7920 B) 74841
C) 14874 D) 10213
 
Answer & Explanation Answer: A) 7920

Explanation:

ST candidates vacancies can be filled by 5C3 ways = 10

Remaining vacancies are 5 that are to be filled by 12

=> 12C5 = (12x11x10x9x8)/(5x4x3x2x1) = 792

Total number of filling the vacancies = 10 x 792 = 7920

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9 297
Q:

A,B,C,D,E,F,G and H are sitting around a circular table facing the centre but not necessarily in the same order. G sits third to the right of C. E is second to the right of G and 4th to the right of H. B is fourth to the right of C. D is not an immediate neighbour of E. A and C are immediate neighbours.

 Which of the following is/are correct ?

A) F is third to the left of B B) F is second to the right of B
C) B is an immediate neighbour of D D) All of the above
 
Answer & Explanation Answer: B) F is second to the right of B

Explanation:

From the given information, the circular arrangement is

EXP.

Here F is second to the right of B and the remaning all are wrong.

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5 171
Q:

Nine different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, how many such words can be formed which have at least one letter repeated ?

A) 43929 B) 59049
C) 15120 D) 0
 
Answer & Explanation Answer: A) 43929

Explanation:

Number of words with 5 letters from given 9 alphabets formed =

Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is = 

Number of words can be formed which have at least one letter repeated =  -  

= 59049 - 15120

= 43929

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8 301
Q:

In how many different ways can the letters of the word 'ABYSMAL' be arranged ?

A) 5040 B) 3650
C) 4150 D) 2520
 
Answer & Explanation Answer: D) 2520

Explanation:

Total number of letters in the word ABYSMAL are 7

Number of ways these 7 letters can be arranged are 7! ways

But the letter is repeated and this can be arranged in 2! ways

=> Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.

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11 456
Q:

In how many ways the letters of the word OLIVER be arranged so that the vowels in the word always occur in the dictionary order as we move from left to right ?

A) 186 B) 144
C) 136 D) 120
 
Answer & Explanation Answer: D) 120

Explanation:

In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O.

There are 6 letters in thegiven word.

First arrange 3 vowels.

This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O)

Remaining 3 letters can be placed in 3 places = 3! ways

Total number of possible ways of arranging letters of OLIVER = 3! x 6C3 ways

= 6x5x4 = 120 ways.

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8 594