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Q:

# From a deck of 52 cards, a 5 card hand is dealt.How may distinct five card hands are there if the queen of spades and the four of diamonds must be in the hand?

 A) 52C5 B) 50C3 C) 52C4 D) 50C4

Explanation:

If the queen of spades and the four of diamonds must be in hand,we have 50 cards remaining out of which we are choosing 3.

So, $\inline&space;{\color{Blue}&space;50C_{3}}$

Q:

Find the total numbers greater than 4000 that can be formed with digits 2, 3, 4, 5, 6 no digit being repeated in any number ?

 A) 120 B) 256 C) 192 D) 244

Explanation:

We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated.

The number can be 4 digited but greater than 4000 or 5 digited.

Number of 4 digited numbers greater than 4000 are
4 or 5 or 6 can occupy thousand place => 3 x 4P3 = 3 x 24 = 72.

5 digited numbers = 5P5 = 5! = 120

So the total numbers greater than 4000 = 72 + 120 = 192

7 105
Q:

In how many ways the letters of the word NUMERICAL can be arranged so that the consonants always occupy the even places ?

 A) 5! B) 6! C) 4! D) Can't determine

Explanation:

NUMERICAL has 9 positions in which 2, 4, 6, 8 are even positions.

And it contains 5 consonents i.e, N, M, R, C & L. Hence this cannot be done as 5 letters cannot be placed in 4 positions.

Therefore, Can't be determined.

5 88
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.

7 214
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.

7 170
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.