8
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

Explanation:

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

$\inline \therefore$ The number of ways in which 9 letters can be arranged = $\inline \frac{9!}{2!\times 2!\times 2!}$ = 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 $\inline \frac{6!}{2!\times 2!}$ = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in $\inline \frac{4!}{2!}$ = 12 ways.

$\inline \therefore$ 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:

36 identical books must be arranged in rows with the same number of books in each row. Each row must contain at least three books and there must be at least three rows. A row is parallel to the front of the room. How many different arrangements are possible ?

 A) 5 B) 6 C) 7 D) 8

Explanation:

The following arrangements satisfy all 3 conditions.

Arrangement 1: 3 books in a row; 12 rows.

Arrangement 2: 4 books in a row; 9 rows.
Arrangement 3: 6 books in a row; 6 rows.
Arrangement 4: 9 books in a row; 4 rows.
Arrangement 5: 12 books in a row; 3 rows.

Therefore, the possible arrangements are 5.

6 75
Q:

There are 11 True or False questions. How many ways can these be answered ?

 A) 11!/2 B) 1024 C) 11! D) 2048

Explanation:

Given 11 questions of type True or False

Then, Each of these questions can be answered in 2 ways (True or false)

Therefore, no. of ways of answering 11 questions = $\inline \fn_jvn 2^{11}$ = 2048 ways.

7 46
Q:

A class has 8 football players. A 5-member team and a captain will be selected out of these 8 players. How many different selections can be made ?

 A) 210 B) 168 C) 1260 D) 10!/6!

Explanation:

we can select the 5 member team out of the 8 in 8C5 ways = 56 ways.
The captain can be selected from amongst the remaining 3 players in 3 ways.
Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways.

(or)

Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 28x6 = 168 ways.

7 43
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.

1 155
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