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Q:

In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

A) 36	B) 25
C) 42	D) 120

Answer & Explanation Answer: A) 36

Explanation:

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = $3 P_{3}$ = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = $3 P_{3}$ = 3! = 6.

Total number of ways = (6 x 6) = 36.

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Q:

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

A) 48	B) 64
C) 63	D) 45

Answer & Explanation Answer: B) 64

Explanation:

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways= $(3 C_{1} * 6 C_{2}) + (3 C_{2} * 6 C_{1}) + 3 C_{3}$ = (45 + 18 + 1) =64

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Q:

The PS command prints the process status for only some of the running processes.

A) TRUE

B) FALSE

Answer & Explanation Answer: B) FALSE

Explanation:

The PS command prints the process status for some or all of the running processes

Workspace

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Subject: Operating Systems
Exam Prep: GATE

Q:

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

A) 209	B) 290
C) 200	D) 208

Answer & Explanation Answer: A) 209

Explanation:

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways = $(6 C_{1} * 4 C_{3}) + (6 C_{2} * 4 C_{2}) + (6 C_{3} * 4 C_{1}) + 6 C_{4}$

= $(6 C_{1} * 4 C_{1}) + (6 C_{2} * 4 C_{2}) + (6 C_{3} * 4 C_{1}) + 6 C_{2}$ = 209.

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Q:

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

A) 25200	B) 52000
C) 120	D) 24400

Answer & Explanation Answer: A) 25200

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = ( $7 C_{3}$ * $4 C_{2}$ )

= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = 120

Required number of ways = (210 x 120) = 25200.

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Q:

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

A) 720	B) 520
C) 700	D) 750

Answer & Explanation Answer: A) 720

Explanation:

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

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Q:

Which among the following is not aprocess state in unix?

A) Running	B) Runnable
C) Zombie	D) Stopped

Answer & Explanation Answer: B) Runnable

Explanation:

The process states of unix are running,Zombie,stopped,Waiting.

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Filed Under: Operating Systems
Exam Prep: GATE

Q:

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

A) 564	B) 735
C) 756	D) 657

Answer & Explanation Answer: C) 756

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways= $(7 C_{3} * 6 C_{2}) + (7 C_{4} * 6 C_{1}) + 7 C_{5}$ = 756.

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Filed Under: Permutations and Combinations

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