How many words can be formed from the letters of the word "SIGNATURE" so that vowels always come together.

A. 17280

B. 4320

C. 720

D. 80

17280

word SIGNATURE contains total 9 letters.

There are four vowels in this word, I, A, U and E

Make it as, SGNTR(IAUE), consider all vowels as 1 letter for now

So total letter are 6.

6 letters can be arranged in 6! ways = 720 ways

Vowels can be arranged in themselves in 4! ways = 24 ways

Required number of ways = 720*24 = 17280

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. 645

C. 735

D. 756

756

Committee may have (3 men and 2 women) or (4 men and 1 women) or (5 men)

Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)

As nCr = nC(n-r), then required number of ways can be written as

= (7C3 x 6C2) + (7C3 x 6C1) + (7C2)

= (525 + 210 + 21)

= 756

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. 960

720

The word LEADING has 7 different letters.

If the vowels E, A, I always come together, then they are considered as one letter.

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

Now, 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

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

A. 10400

B. 30400

C. 50400

D. 70400

50400

In the word CORPORATION, we consider the vowels OOAIO as one letter.

Then, we have CRPRTN (OOAIO).

Now we have 7 letters of which R occurs 2 times.

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

Now, in 5 vowels in which O occurs 3 times

Number of ways arranging vowels letters = 5! / 3! = 20 ways

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

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

A. 5200

B. 15200

C. 25200

D. 35200

25200

Number of ways of selecting (3 consonants from 7) and (2 vowels from 4)

= 7C3 x 4C2

= 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

In how many ways can the letters of the word LEADER be arranged?

A. 90

B. 180

C. 360

D. 720

360

The word LEADER contains 1L, 2E, 1A, 1D and 1R

=> Required number of ways = 6! / (1!) (2!) (1!) (1!) (1!) = 360

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. 160

B. 189

C. 205

D. 209

209

The required number of ways 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 = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)

As nCr = nC(n-r) we can write above equation as

= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)

= (24 + 90 + 80 + 15)

= 209

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. 5

B. 10

C. 15

D. 20

20

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 ways = (1 x 5 x 4) = 20

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

A. 560

B. 5040

C. 11760

D. 86400

11760

Required number of ways = (8C5 x 10C6)

As nCr = nC(n-r) we can write above equation as

= (8C3 x 10C4)

= 11760

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. 16

B. 32

C. 64

D. 128

64

The required number of ways can be formed as (1 black and 2 non-black) or (2 black and 1 non-black) or (3 black)

=> Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)

= (45 + 18 + 1)

= 64

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. 72

B. 36

C. 18

D. 9

36

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 ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

A. 63

B. 126

C. 45

D. 90

63

Required number of ways = (7C5 x 3C2)

As nCr = nC(n-r) we can write as

= (7C2 x 3C1)

= 63

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. 1250

B. 3540

C. 5040

D. 2520

5040

LOGARITHMS contains 10 different letters.

Required number of words = 10P4

= 5040

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. 350609

120960

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!) / (2!) (2!) = 10080

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

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

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

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

A. 360

B. 720

C. 1440

D. 2880

720

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

In how many ways can the letters of the word 'LEADER' be arranged?

A. None of these

B. 120

C. 360

D. 720

360

No Explanation available for this question

A local delivery company has three packages to deliver to three different homes. if the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package?

A. 3

B. 5

C. 3!

D. 5!

5

No Explanation available for this question

How many total numbers of not more than 20 digits that can be formed by using the digits 0 ,1 ,2 ,3 , and 4 is

A. 520

B. 520-1

C. 520+1

D. 620

E. None of these

5^{20}

No Explanation available for this question

How many words with or without meaning, can be formed by using all the letters of the word. “DELHI’ using all the letters of the word ‘DELHI’ using each letter exactly once?

A. 10

B. 25

C. 60

D. 120

120

No Explanation available for this question

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

A. 144

B. 24

C. 6

D. 132

144

No Explanation available for this 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

5040

No Explanation available for this question

In how many ways can 6 green toys and 6 red toys be arranged, such that 2 particular red toys are never together whereas 2 particular green toys are always together?

A. 11! × 2!

B. 9! × 90

C. 4 × 10!

D. 18 × 10!

18 × 10!

No Explanation available for this question

In how many ways can the letters of the word ’RUMOUR’ be arranged

A. 180

B. 90

C. 30

D. 720

180

No Explanation available for this question

There are 6 boxes numbered 1,2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:

A. 5

B. 21

C. 33

D. 60

E. 27

21

No Explanation available for this question

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

A. 40

B. 400

C. 5040

D. 2520

5040

No Explanation available for this question

An urn contains 5 red, 3 green, 2 blue and 4 yellow marbles. If three marbles are picked at random, what is the probability that at least one is yellow?

A. 2/7

B. 4/91

C. 27/143

D. 61/91

E. None of these

61/91

No Explanation available for this question

A teacher of 6 students takes 2 of his students at a time to a zoo as often as he can, without taking the same pair of children together more than once. How many times does the teacher go to the zoo?

A. 10

B. 12

C. 15

D. 20

15

No Explanation available for this question

How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3, and 4, if repetition of digits is allowed?

A. 499

B. 500

C. 375

D. 376

E. 501

376

No Explanation available for this question

How many five digits positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4, and 5, without any of the digits getting repeating?

A. 15

B. 96

C. 216

D. 120

E. 625

216

No Explanation available for this question

There are 8 Orange, 6 White and 7 Black pens in a Parul Bag. Parul picked up randomly a pen. What is the probability that it is neither orange nor white?

A. 1/6

B. 7/3

C. 21/14

D. 1/3

1/3

No Explanation available for this question