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Q:

Find the number of ways to take 20 objects and arrange them in groups of 5 at a time where order does not matter.?

A) 57090	B) 15540
C) 15504	D) 23670

Answer & Explanation Answer: C) 15504

Explanation:

${}^{20}C_{5}$ = 15504

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Q:

Find the number of ways to take 4 people and place them in groups of 3 at a time where order does not matter?

A) 4	B) 12
C) 36	D) 16

Answer & Explanation Answer: A) 4

Explanation:

Since order does not matter, use the combination formula

${}^{4}C_{3}$ = 24/6 = 4

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Q:

Find the number of ways to arrange 6 items in groups of 4 at a time where order matters?

A) 720	B) 640
C) 740	D) 360

Answer & Explanation Answer: D) 360

Explanation:

6P4 = 6! / (6-4)! = 360

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Q:

Find the number of ways to arrange 4 people in groups of 3 at a time where order matters?

A) 20	B) 16
C) 24	D) 36

Answer & Explanation Answer: C) 24

Explanation:

P(4,3)= ${}^{4}P_{3}$ = 24

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Q:

If repetition of the digits is allowed, then the number of even natural numbers having three digits is :

A) 550	B) 450
C) 500	D) 540

Answer & Explanation Answer: B) 450

Explanation:

In a 3 digit number one’s place can be filled in 5 different ways with (0,2,4,6,8)

10’s place can be filled in 10 different ways

100’s place can be filled in 9 different ways

There fore total number of ways = 5X10X9 = 450

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Q:

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :

A) 601	B) 600
C) 603	D) 602

Answer & Explanation Answer: A) 601

Explanation:

If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.

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Q:

Find the total number of distinct vehicle numbers that can be formed using two letters followed by two numbers. Letters need to be distinct

A) 65000	B) 64000
C) 72000	D) 36000

Answer & Explanation Answer: A) 65000

Explanation:

Out of 26 alphabets two distinct letters can be chosen in $26 P_{2}$ ways. Coming to numbers part, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways.

Combined with letters there are $6 P_{2}$ X 100 ways = 65000 ways to choose vehicle numbers.

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Q:

How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?

A) 1000	B) 100
C) 500	D) 999

Answer & Explanation Answer: B) 100

Explanation:

1 million distinct 3 digit initials are needed.

Let the number of required alphabets in the language be ‘n’.

Therefore, using ‘n’ alphabets we can form n * n * n = $n^{3}$ distinct 3 digit initials.

Note distinct initials is different from initials where the digits are different.

For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.

This $n^{3}$ different initials = 1 million

i.e. $n^{3} = 10^{6}$ (1 million = $10^{6}$ )

=> n = $10^{2}$ = 100

Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.

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