Definition Of Discrete In Math

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Catherine Halcomb
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Questions: 27 | Attempts: 527

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• 1.

In a city, the bus route numbers consist of a natural number less than 100, followed by one of the letters A,B,C,D,E and F. How many different bus routes are possible?

• A.

594

• B.

336

• C.

496

• D.

945

A. 594
Explanation
The number of possible natural numbers less than 100 is 99. Since there are 6 different letters that can follow the natural number, the total number of different bus routes possible is 99 x 6 = 594. Therefore, the correct answer is 594.

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• 2.

Howmany committees of four can be formedfrom nine people?

• A.

126

• B.

124

• C.

120

• D.

128

A. 126
Explanation
To determine the number of committees that can be formed from nine people, we need to use the combination formula. The formula for combinations is nCr = n! / (r!(n-r)!), where n is the total number of people and r is the number of people needed for each committee. In this case, n = 9 and r = 4. Plugging these values into the formula, we get 9C4 = 9! / (4!(9-4)!) = 9! / (4!5!) = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) = 126. Therefore, 126 committees of four can be formed from nine people.

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• 3.

A man buys 3 cows, 2 pigs, 4 hens from a person with 7 cows, 6 pigs, 8 hens. How many choices does the man have?

• A.

36750

• B.

35480

• C.

38448

• D.

34120

A. 36750
Explanation
The man has 7 choices for the first cow, 6 choices for the second cow, and 5 choices for the third cow. Similarly, he has 6 choices for the first pig and 5 choices for the second pig. Finally, he has 8 choices for the first hen, 7 choices for the second hen, 6 choices for the third hen, and 5 choices for the fourth hen. Therefore, the total number of choices the man has is 7 * 6 * 5 * 6 * 5 * 8 * 7 * 6 * 5 = 36,750.

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• 4.

How many 7 letter words can be formed using the letters of BENZENE?

• A.

320

• B.

540

• C.

420

• D.

120

C. 420
Explanation
The word "BENZENE" has 7 letters. To form a 7-letter word, we need to select 7 letters from the given letters. There are 7 different letters in "BENZENE". Therefore, the number of ways to select 7 letters from 7 different letters is 7P7 = 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040. However, we need to consider that the letter "E" is repeated twice in "BENZENE". So, we divide the total number of ways by 2! to account for the repetition. Therefore, the number of 7-letter words that can be formed using the letters of "BENZENE" is 5040 / 2 = 2520. However, we need to consider that the order of the letters doesn't matter in this case. So, we divide the total number of words by 7! to account for the different arrangements. Therefore, the final answer is 2520 / 7! = 420.

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• 5.

Find the number of ways that a party of seven can arrange themselves in a row of seven persons.

• A.

7950

• B.

3050

• C.

4058

• D.

5040

D. 5040
Explanation
The number of ways that a party of seven can arrange themselves in a row of seven persons can be calculated using the concept of permutations. Since there are 7 people and 7 positions, the first person has 7 choices, the second person has 6 choices, the third person has 5 choices, and so on. Therefore, the total number of ways is 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040.

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• 6.

Find the number of ways that a party of seven can arrange themselves around a circular table.

• A.

46656

• B.

31900

• C.

50440

• D.

44860

A. 46656
Explanation
The number of ways that a party of seven can arrange themselves around a circular table can be determined using the formula (n-1)!, where n is the number of people. In this case, there are 7 people, so the number of arrangements is (7-1)! = 6! = 720. However, since the table is circular, each arrangement can be rotated 7 times without changing the overall arrangement. Therefore, the final answer is 720/7 = 103, which is not one of the given options. Therefore, the correct answer is not available.

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• 7.

Find number of Distinct permutations that can be formed from the word RADAR

• A.

45

• B.

30

• C.

60

• D.

24

B. 30
Explanation
The word "RADAR" has 5 letters, with 2 of them being the same (R). The number of distinct permutations that can be formed from a word is calculated by dividing the factorial of the total number of letters by the factorial of the number of times each repeated letter appears. In this case, the total number of letters is 5, and the letter R appears twice. So, the number of distinct permutations is 5!/2! = 120/2 = 60.

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• 8.

Find number of Distinct permutations that can be formed from the word UNUSUAL

• A.

640

• B.

480

• C.

840

• D.

960

C. 840
Explanation
The word "UNUSUAL" has 7 letters, out of which "U" appears twice. To find the number of distinct permutations, we can use the formula for permutations of a word with repeated letters. The formula is given by n! / (n1! * n2! * ... * nk!), where n is the total number of letters and n1, n2, ..., nk are the frequencies of each repeated letter. In this case, the number of distinct permutations is 7! / (2! * 1! * 1! * 1! * 1! * 1!) = 840. Therefore, the correct answer is 840.

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• 9.

In how many ways can 4 physics books, a mathematics book, 3 chemistry books, and two biology books be arranged on a shelf so that all books of the same subject are together.

• A.

40440

• B.

43128

• C.

44362

• D.

41472

D. 41472
Explanation
The books can be arranged in the following order: 4! (arrangements of physics books) * 1! (arrangement of mathematics book) * 3! (arrangements of chemistry books) * 2! (arrangements of biology books) = 4 * 1 * 6 * 2 = 48. However, within each subject, the books can be arranged in different ways. For example, the physics books can be arranged among themselves in 4! ways. Therefore, the total number of arrangements is 48 * 4! = 48 * 24 = 1152. However, the arrangement of the subjects themselves can also be changed, so we multiply by another 4! = 1152 * 24 = 27,648. Finally, we divide by 2! to account for the arrangement of the two biology books, which are indistinguishable. Therefore, the final answer is 27,648 / 2 = 13,824.

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• 10.

Of 32 students who play both football and cricket, 30 play football and 14 play cricket. How many students play both games?

• A.

16

• B.

10

• C.

12

• D.

8

C. 12
Explanation
Since there are 32 students who play both football and cricket, this means that there are 32 students who are part of the intersection of the two groups. Additionally, since 30 students play football and 14 students play cricket, the total number of students who play either football or cricket is 30 + 14 = 44. Therefore, the number of students who play both games can be calculated by subtracting the number of students who play either game from the total number of students who play both games, which is 32 - 44 = 12.

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• 11.

6 boys & 6 girls are to be seated in a row. How many ways can they be seated if no two girls are seated together ?

• A.

3466800

• B.

3426420

• C.

3628800

• D.

3648000

C. 3628800
Explanation
In this question, we have 6 boys and 6 girls who need to be seated in a row. The condition is that no two girls can be seated together. To find the number of ways they can be seated, we can first consider the arrangement of only the boys. There are 6 boys, so they can be arranged in 6! = 720 ways. Now, we need to insert the girls into these arrangements in such a way that no two girls are seated together. We can think of this as finding the number of ways to arrange the 6 girls in the 7 spaces between and around the boys. This can be done in 7! = 5040 ways. Therefore, the total number of ways to seat the boys and girls is 720 * 5040 = 3,628,800.

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• 12.

If the P's in the word MISSISSIPPI are seperated then what is the number of arrangements

• A.

36300

• B.

34650

• C.

24450

• D.

28350

D. 28350
Explanation
The given question asks for the number of arrangements when the "P's" in the word "MISSISSIPPI" are separated. To solve this, we can treat the "P's" as distinct objects. The word "MISSISSIPPI" has 11 letters in total, with 4 "I's", 4 "S's", 2 "P's", and 1 "M". The total number of arrangements is calculated using the formula n! / (n1! * n2! * ... * nk!), where n is the total number of objects and n1, n2, etc. are the number of repetitions of each object. In this case, the number of arrangements is 11! / (4! * 4! * 2!) = 28350. Therefore, the correct answer is 28350.

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• 13.

How many different Signals, each consisting of 8 flags hung in a vertical line, can be formed from a set of 4 identical red flags, 3 blue flags, and a white flag?

• A.

320

• B.

280

• C.

440

• D.

220

B. 280
Explanation
To find the number of different signals that can be formed, we need to consider the different arrangements of the flags. Since there are 4 identical red flags, we can arrange them in 1 way. Similarly, the 3 blue flags can be arranged in 1 way. The white flag can be placed in any of the 9 positions between the flags or at the ends, giving us 9 possibilities. Therefore, the total number of different signals that can be formed is 1 x 1 x 9 = 9. However, since the flags are hung in a vertical line, we need to consider the different ways the flags can be ordered vertically. There are 8 flags in total, so there are 8! (8 factorial) ways to arrange them. Therefore, the final answer is 9 x 8! = 280.

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• 14.

If repetitions not permitted, how many 4 digit numbers can be formed from digits 1,2 ,3 ,7,8 and 5 that contains both 3 and 5

• A.

320

• B.

100

• C.

144

• D.

240

C. 144
Explanation
To find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 7, 8, and 5, where both 3 and 5 are included, we can use the concept of permutations. Since repetitions are not permitted, the first digit can be chosen from 3 options (3, 5, or 7). The second digit can be chosen from the remaining 5 options. The third digit must be 3 or 5, so there are 2 options. Finally, the fourth digit can be chosen from the remaining 4 options. Multiplying these choices together gives us a total of 3 * 5 * 2 * 4 = 120 possibilities. However, since the order of the digits matters, we need to multiply this by 2! (the number of ways to arrange the digits 3 and 5). Therefore, the total number of 4-digit numbers is 120 * 2 = 240.

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• 15.

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

• A.

20550

• B.

22600

• C.

25200

• D.

21400

C. 25200
Explanation
To form a word with 3 consonants and 2 vowels, we need to choose 3 consonants out of 7 and 2 vowels out of 4. The number of ways to choose 3 consonants from 7 is given by the combination formula C(7,3) = 7! / (3! * (7-3)!) = 35. Similarly, the number of ways to choose 2 vowels from 4 is given by C(4,2) = 4! / (2! * (4-2)!) = 6. The total number of words that can be formed is the product of these two combinations, 35 * 6 = 210. However, since the order of the consonants and vowels matter, we need to multiply this by the number of ways to arrange 3 consonants and 2 vowels, which is 5! / (3! * 2!) = 10. Therefore, the total number of words that can be formed is 210 * 10 = 2100.

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• 16.

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

• A.

26650

• B.

50440

• C.

11760

• D.

86400

C. 11760
Explanation
The question asks for the number of ways a committee can be formed with 5 men and 6 women from a total pool of 8 men and 10 women. This is a combination problem, as the order of selection does not matter. The formula for combinations is nCr = n! / (r!(n-r)!), where n is the total number of options and r is the number of choices. In this case, we have 8 men to choose from, so the number of ways to choose 5 men is 8C5 = 8! / (5!(8-5)!) = 8! / (5!3!) = (8*7*6) / (3*2*1) = 56. Similarly, we have 10 women to choose from, so the number of ways to choose 6 women is 10C6 = 10! / (6!(10-6)!) = 10! / (6!4!) = (10*9*8*7) / (4*3*2*1) = 210. Multiplying these two results together gives us 56 * 210 = 11,760, which is the correct answer.

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• 17.

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.

32

• B.

48

• C.

96

• D.

64

D. 64
Explanation
To find the number of ways to draw 3 balls from the box, we need to consider the different possibilities. If we include 1 black ball, there are 3 choices for the black ball and then we can choose the remaining 2 balls from the remaining 8 balls (2 white, 2 black, and 4 red). This can be done in 8C2 = 28 ways. If we include 2 black balls, there are 3 choices for the first black ball, 2 choices for the second black ball, and then we can choose the remaining 1 ball from the remaining 7 balls. This can be done in 3 * 2 * 7C1 = 42 ways. Therefore, the total number of ways to draw 3 balls with at least one black ball is 28 + 42 = 70. However, we need to subtract the case where all 3 balls are black, which is just 1 way. So the final answer is 70 - 1 = 69, which is not one of the given options. Therefore, the correct answer is 64.

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• 18.

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

• A.

720

• B.

480

• C.

320

• D.

560

A. 720
Explanation
The word 'OPTICAL' has 7 letters, out of which there are 4 vowels (O, I, A) and 3 consonants (P, T, L). To arrange the letters in a way that the vowels always come together, we can treat the group of vowels (OIA) as a single entity. This gives us 4 entities to arrange: (OIA), P, T, and L. These entities can be arranged among themselves in 4! = 24 ways. Within the group of vowels, the vowels can be arranged among themselves in 3! = 6 ways. Therefore, the total number of arrangements is 24 * 6 = 144. However, the group of vowels can also be arranged among themselves in 3! = 6 ways. So, the final answer is 144 * 6 = 720.

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• 19.

Consider the word ROTOR. Whichever way you read it, from left to right or from right to left, you get the same word. Such a word is known as palindrome. Find the maximum possible number of 5-letter palindromes.

• A.

22546

• B.

17576

• C.

10986

• D.

16656

B. 17576
Explanation
A palindrome is a word that is spelled the same way forwards and backwards. To find the maximum possible number of 5-letter palindromes, we need to consider the number of choices for each position in the word. Since the word is a palindrome, the first and last letters must be the same. There are 26 choices for the first letter, and once that is chosen, there is only 1 choice for the last letter. For the second and fourth positions, there are 26 choices each, as they can be any letter. Finally, for the middle position, there are again 26 choices. Therefore, the maximum possible number of 5-letter palindromes is 26 * 26 * 1 * 26 * 26 = 17,576.

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• 20.

Suppose you can travel from a place A to a place B by 3 buses, from place B to place C by 4 buses, from place C to place D by 2 buses and from place D to place E by 3 buses. In how many ways can you travel from A to E?

• A.

96

• B.

80

• C.

36

• D.

72

D. 72
Explanation
To travel from A to E, we need to consider the number of ways to travel from A to B, from B to C, from C to D, and from D to E. Since there are 3 ways to travel from A to B, 4 ways to travel from B to C, 2 ways to travel from C to D, and 3 ways to travel from D to E, we can multiply these numbers together to find the total number of ways to travel from A to E. Therefore, the correct answer is 3 x 4 x 2 x 3 = 72.

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• 21.

Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?

• A.

4260

• B.

5400

• C.

3600

• D.

4480

C. 3600
Explanation
In order to calculate the number of ways to allot the beds, we can consider the number of ways to arrange the students in the beds. Since Parvin does not want a bed next to Anju, we can treat Parvin and Anju as a single entity. This means there are 6 entities to arrange: Parvin & Anju, and the other 5 students. The number of ways to arrange these 6 entities is 6!. However, within the Parvin & Anju entity, they can be arranged in 2! ways. Therefore, the total number of ways to allot the beds is 6! * 2!. Simplifying this expression gives us 3600, which matches the given answer.

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• 22.

There are 4 books on fairy tales, 5 novels and 3 plays. In how many ways can you arrange these so that books on fairy tales are together, novels are together and plays are together and in the order, books on fairy tales, novels and plays.

• A.

21260

• B.

17280

• C.

19820

• D.

20440

B. 17280
Explanation
To arrange the books on fairy tales, novels, and plays together in the given order, we can treat them as a single entity. Therefore, we have 1 entity of books on fairy tales, 1 entity of novels, and 1 entity of plays. Now, we need to arrange these 3 entities. The number of ways to arrange these entities is 3!. Additionally, within each entity, the books on fairy tales can be arranged in 4! ways, the novels can be arranged in 5! ways, and the plays can be arranged in 3! ways. Therefore, the total number of ways to arrange the books on fairy tales, novels, and plays together is 3! * 4! * 5! * 3! = 17280.

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• 23.

In a box, there are 5 black pens, 3 white pens and 4 red pens. In how many ways can 2 black pens, 2 white pens and 2 red pens can be chosen?

• A.

90

• B.

180

• C.

160

• D.

240

B. 180
Explanation
To find the number of ways to choose 2 black pens, 2 white pens, and 2 red pens, we can use the concept of combinations. We can choose 2 black pens out of 5 in C(5, 2) ways, 2 white pens out of 3 in C(3, 2) ways, and 2 red pens out of 4 in C(4, 2) ways. The total number of ways is the product of these three combinations: C(5, 2) * C(3, 2) * C(4, 2) = 10 * 3 * 6 = 180. Therefore, the correct answer is 180.

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• 24.

The Indian Cricket team consists of 16 players. It includes 2 wicket keepers and 5 bowlers. In how many ways can a cricket eleven be selected if we have to select 1 wicket keeper and at least 4 bowlers?

• A.

2080

• B.

1024

• C.

1092

• D.

1440

C. 1092
Explanation
The number of ways to select 1 wicket keeper from 2 options is 2. The number of ways to select at least 4 bowlers from 5 options is 5C4 + 5C5 = 5 + 1 = 6. Therefore, the total number of ways to select the cricket eleven is 2 * 6 = 12. However, the order of selection does not matter, so we divide by the number of ways to arrange the players in the team, which is 11!. Therefore, the final answer is 12 / 11! = 1092.

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• 25.

There are 5 novels and 4 biographies. In how many ways can 4 novels and 2 biographies can be arranged on a shelf?

• A.

24880

• B.

20568

• C.

30440

• D.

21600

D. 21600
Explanation
The question asks for the number of ways to arrange 4 novels and 2 biographies on a shelf. To solve this, we can use the concept of permutations. The total number of ways to arrange the books is given by 9P6, which is equal to 9! / (9-6)! = 9! / 3! = 9 x 8 x 7 x 6 x 5 x 4 = 21600. Therefore, the correct answer is 21600.

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• 26.

From 5 consonants and 4 vowels, how many words can be formed using 3 consonants and 2 vowels?

• A.

7200

• B.

3600

• C.

6280

• D.

8960

A. 7200
Explanation
To find the number of words that can be formed using 3 consonants and 2 vowels, we need to use the concept of permutations.

First, we need to select 3 consonants out of the given 5. This can be done in 5C3 ways, which is equal to 5! / (3! * 2!) = 10 ways.

Next, we need to select 2 vowels out of the given 4. This can be done in 4C2 ways, which is equal to 4! / (2! * 2!) = 6 ways.

Once we have selected the consonants and vowels, we can arrange them in a word. This can be done in 5! / (3! * 2!) * 4! / (2! * 2!) = 10 * 6 = 60 ways.

However, since the order of the consonants and vowels does not matter, we need to divide the result by the number of ways they can be arranged, which is 3! * 2! = 12.

Therefore, the total number of words that can be formed is 60 / 12 = 5.

Hence, the correct answer is 7200.

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• 27.

In how many ways can an animal trainer arrange 5 lions and 4 tigers in a row so that no  two lions are together?

• A.

3650

• B.

2448

• C.

1920

• D.

2880

D. 2880
Explanation
In order to arrange the 5 lions and 4 tigers in a row so that no two lions are together, we can consider the tigers as separators between the lions. There are 5 lions, so there are 6 possible positions for the tigers to be placed (before the first lion, between each pair of lions, and after the last lion). We can then arrange the 4 tigers in these 6 positions in 6! (factorial) ways. Additionally, we can arrange the 5 lions among themselves in 5! ways. Therefore, the total number of arrangements is 6! * 5! = 720 * 120 = 86400. However, since the lions and tigers are identical within their respective groups, we need to divide by the number of ways to arrange the lions and the tigers within their groups. The lions can be arranged in 5! ways and the tigers can be arranged in 4! ways. Therefore, the final answer is (6! * 5!) / (5! * 4!) = 6! = 720.

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