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How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?

A.

360

B.

720

C.

1420

D.

1680

Answer with explanation

Answer: Option DExplanation

The first two places can only be filled by 3 and 5 respectively and there is only 1 way for doing this.

Given that no digit appears more than once. Hence we have 8 digits remaining (0,1,2,4,6,7,8,9)(0,1,2,4,6,7,8,9)

So, the next 4 places can be filled with the remaining 8 digits in ^{8}P_{4} ways.

Total number of ways = ^{8}P_{4} =8×7×6×5=1680

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How many different four letter words can be formed (the words need not be meaningful using the letters of the word “MEDITERRANEAN” such that the first letter is E and the last letter is R?

A.

53

B.

55

C.

59

D.

69

Answer with explanation

Answer: Option CExplanation

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters. Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

**Case 1**: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

**Case 2**: When the two letters are same. There are 3 options – the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59

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In how many ways can a committee consisting of three men and four women be formed from a group of six men and seven women?

A.

⁶C₃ ⁷C₅

B.

⁶C₃⁷C₄

C.

⁷C₅ ⁶C₄

D.

⁶C₃ ⁷C₃

Answer with explanation

Answer: Option BExplanation

The group contained six men and seven women

Three men can be selected from six men in ⁶C₃ ways.

Four women can be selected from seven women in ⁷C₄ ways.

A total number of ways = (⁷C₄)(⁶C₃).

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A box contains 2 red coins, 3 green coins and 4 blue coins. In how many ways can 3 coins be chosen such that at least one coin is green?

A.

128

B.

32

C.

16

D.

64

Answer with explanation

Answer: Option DExplanation

There are three cases:

- 3 green coins
- 2 green coins + 1 non-green coin
- 1 green coin + 2 non-green coins

Therefore, total number of ways = ^{3}C_{3} + ^{3}C_{2} * ^{6}C_{1} + ^{3}C_{1} * ^{6}C_{2} = 1 + 3*6 + 3*15 = 64.

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In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?

A.

9!/(4!*5!)

B.

4!*5!

C.

9!/4

D.

None of these

Answer with explanation

Answer: Option BExplanation

The word EDUCATION is a 9 letter word, with none of the letters repeating.

The vowels occupy 3, 5, 7^{th} and 8^{th} position in the word and the remaining 5 positions are occupied by consonants.

As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the aforementioned 4 places and the consonants can occupy 1^{st}, 2^{nd}, 4^{th}, 6^{th} and 9^{th} positions.

The 4 vowels can be arranged in the 3^{rd}, 5^{th}, 7^{th} and 8^{th} position in 4! Ways.

Similarly, the 5 consonants can be arranged in 1^{st}, 2^{nd}, 4^{th}, 6^{th} and 9^{th} position in 5! Ways.

Hence, the total number of ways = 4! * 5!.

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In how many ways can you rearrange the word JUMBLE such that the rearranged word starts with a vowel?

A.

60

B.

120

C.

240

D.

360

Answer with explanation

Answer: Option CExplanation

JUMBLE is a six-lettered word. Since the rearranged word has to start with a vowel, the first letter can be either U or E. The balance 5 letters can be arranged in ^{5}P_{5} or 5! ways. Total number of words = 2 × 5! = 240.

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How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

A.

20

B.

15

C.

10

D.

5

Answer with explanation

Answer: Option AExplanation

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

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How many arrangements can be made out of the letters of the word ‘ENGINEERING’?

A.

277200

B.

182000

C.

924000

D.

734500

Answer with explanation

Answer: Option AExplanation

The word ‘ENGINEERING’ has 11 letters.

But in these 11 letters, ‘E’ occurs 3 times,’N’ occurs 3 times, ‘G’ occurs 2 times, ‘I’ occurs 2 times and rest of the letters are different.

Hence,number of ways to arrange these letters

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