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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?

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Question: 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].

32

[B].

48

[C].

36

[D].

60

Answer: Option C

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 = 3P3 = 3! = 6.

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

Number of ways of these arrangements = 3P3 = 3! = 6.

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

In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together?

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Question:
In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together?

[A].

10080

[B].

4989600

[C].

120960

[D].

None of these

Answer: Option C

Explanation:

In the word ‘MATHEMATICS’, we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters = 8! = 10080.
(2!)(2!)

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters = 4! = 12.
2!

Required number of words = (10080 x 12) = 120960.

In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?

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Question: In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
[A].

810

[B].

1440

[C].

2880

[D].

50400

Answer: Option D

Explanation:

In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters = 7! = 2520.
2!

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged

in 5! = 20 ways.
3!

Required number of ways = (2520 x 20) = 50400.

Video Explanation: https://youtu.be/o3fwMoB0duw

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

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Question: 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].

360

[B].

480

[C].

720

[D].

5040

Answer: Option C

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 = 5) 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.

Video Explanation: https://youtu.be/WCEF3iW3H2c

In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?

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Question:
In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?

[A].

120

[B].

720

[C].

4320

[D].

2160

Answer: Option B

Explanation:

The word ‘OPTICAL’ contains 7 different letters.

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

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

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

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

How many 4-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?

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Question: How many 4-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?
[A].

40

[B].

400

[C].

5040

[D].

2520

Answer: Option C

Explanation:

‘LOGARITHMS’ contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
= 10P4
= (10 x 9 x 8 x 7)
= 5040.

Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

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Question:
Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

[A].

39, 30

[B].

41, 32

[C].

42, 33

[D].

43, 34

Answer: Option C

Explanation:

Let their marks be (x + 9) and x.

Then, x + 9 = 56 (x + 9 + x)
100

25(x + 9) = 14(2x + 9)

3x = 99

x = 33

So, their marks are 42 and 33.