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Q:

How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such that the four vowels do not come together?

A) 216 B) 45360
C) 1260 D) 43200

Answer:   D) 43200



Explanation:

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

 The number of ways in which 9 letters can be arranged =  = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in  = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in  = 12 ways.

 The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

Q:

How many four digit even numbers can be formed using the digits {2, 7, 5, 3, 9, 1} ?

A) 59 B) 60
C) 61 D) 64
 
Answer & Explanation Answer: B) 60

Explanation:

The given digits are 1, 2, 3, 5, 7, 9 

A number is even when its units digit is even. Of the given digits, two is the only even digit.
Units place is filled with only '2' and the remaining three places can be filled in ⁵P₃ ways.
Number of even numbers = ⁵P₃ = 60.
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1 43
Q:

A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are picked at random, what is the probability that they are either blue or yellow ?

A) 3/17 B) 4/21
C) 2/21 D) 5/17
 
Answer & Explanation Answer: C) 2/21

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 x 2)/(15 x 14) = 1/35
Probability that both are yellow = ²C₂/¹⁵C₂ = (2 x 1)/(15 x 14) = 1/105
Probability that one blue and other is yellow = (³C₁ x ²C₁)/¹⁵C₂ = (2 x 3 x 2)/(15 x 14) = 2/35

Required probability = 1/35 + 1/105 + 2/35
= 3/35 + 1/105 = 1/35(3 + 1/3)
= 10/(3 x 35) = 2/21

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4 117
Q:

A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group ?

A) 600 B) 610
C) 609 D) 599
 
Answer & Explanation Answer: A) 600

Explanation:

Two men, three women and one child can be selected in ⁴C₂ x ⁶C₃ x ⁵C₁ ways

=

= 600 ways.

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1 139
Q:

The number of sequences in which 4 players can sing a song, so that the youngest player may not be the last is ?

A) 2580 B) 3687
C) 4320 D) 5460
 
Answer & Explanation Answer: C) 4320

Explanation:

Let 'Y' be the youngest player.

The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways.

The required number of ways = 3(3!) = 4320.

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3 144
Q:

A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most  ?

A) 215 B) 268
C) 254 D) 216
 
Answer & Explanation Answer: A) 215

Explanation:

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts, one of them is a successful attempt.

Maximum number of unsuccessful attempts = 216 - 1 = 215.

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