1
Q:

# In a G - 20 meeting there were total 20 people representing their own country. All the representative sat around a circular table. Find the number of ways in which we can arrange them around a circular table so that there is exactly one person between two representatives namely Manmohan and Musharraf.

 A) 2 x (17!) B) 2 x (18!) C) (3!) x (18!) D) (17!)

Explanation:

A person can be choosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf , Manmohan and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways.

$\inline \therefore$ Required number of permutations = 18 x (17!) x 2 = 2 x 18!

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

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.

1 51
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

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

4 123
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

Explanation:

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

= $\inline \fn_jvn \small \left ( \frac{4x3}{2x1} \right )x \left ( \frac{6x5x4}{3x2x1} \right )x 5$

= 600 ways.

1 144
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

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.

3 150
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