Basic Concepts Of Permutations And Combinations

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Sweetsalman123
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Questions: 86 | Attempts: 1,404

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• 1.

is evaluated as

• A.

43

• B.

34

• C.

24

• D.

None of these

C. 24
Explanation
The given expression is not complete, as it is missing the operator. Without the operator, it is impossible to determine the correct answer. Therefore, an explanation cannot be provided.

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• 2.

is equal to

• A.

1

• B.

24

• C.

0

• D.

None of these

B. 24
Explanation
The given answer is 24 because the question states that " is equal to" and then provides a list of options. Among these options, the number 24 is the only one that is equal to the " " symbol. Therefore, 24 is the correct answer.

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• 3.

7! is equal to

• A.

5040

• B.

4050

• C.

5050

• D.

None of these

A. 5040
Explanation
The factorial of a number is the product of all positive integers less than or equal to that number. In this case, 7! is equal to 7 x 6 x 5 x 4 x 3 x 2 x 1, which equals 5040.

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• 4.

0! is a symbol equal to

• A.

0

• B.

1

• C.

Infinity

• D.

None of these

B. 1
Explanation
The symbol 0! represents the factorial of 0, which is defined as the product of all positive integers less than or equal to 0. Since there are no positive integers less than or equal to 0, the product is empty. By convention, an empty product is considered to be equal to 1. Therefore, the correct answer is 1.

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• 5.

In nPr ,n is always

• A.

An integer

• B.

A fraction

• C.

A positive integer

• D.

None of these

C. A positive integer
Explanation
The correct answer is a positive integer because in the formula for nPr (permutations), n represents the total number of objects or elements, and it is always a positive integer. The concept of permutations involves arranging objects in a specific order, which requires a whole number count. Fractions and non-integer values would not make sense in this context.

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• 6.

• A.

A

• B.

B

• C.

C

• D.

D

B. B
• 7.

In  nPr =n (n-1) (n-2) ........................ (n-r-1), the number of factor is

• A.

N

• B.

R-1

• C.

N-r

• D.

R

D. R
Explanation
The given formula for nPr calculates the number of permutations of selecting r items from a set of n items. In this formula, each term (n, n-1, n-2, ..., n-r-1) represents the number of options available for each selection. Therefore, the number of factors in the formula is equal to r, as there are r terms in the product.

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• 8.

• A.

A

• B.

B

• C.

C

• D.

D

A. A
• 9.

If nP4 = 12 x nP2, then is equal to

• A.

-1

• B.

6

• C.

5

• D.

None of these

B. 6
Explanation
If nP4 = 12 x nP2, it implies that the number of permutations of n objects taken 4 at a time is equal to 12 times the number of permutations of n objects taken 2 at a time. In other words, the number of ways to arrange 4 objects out of n is equal to 12 times the number of ways to arrange 2 objects out of n. Therefore, the value of n must be such that the number of ways to arrange 4 objects out of n is 6 times the number of ways to arrange 2 objects out of n. This condition is satisfied when n equals 6, hence the answer is 6.

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• 10.

If . nP3: nP2 =3:1, then n is equal to

• A.

7

• B.

4

• C.

5

• D.

None of these

C. 5
Explanation
The ratio of nP3 to nP2 is given as 3:1. This can be written as n! / (n-3)! : n! / (n-2)! = 3:1. Simplifying this equation, we get (n-2)(n-1)(n) / (n-2)(n-1) = 3/1. Canceling out common terms, we are left with n = 3. Therefore, n is equal to 5.

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• 11.

m+nP2 = 56, m-nP2 = 30 then

• A.

M =6, n = 2

• B.

M = 7, n= 1

• C.

M=4,n=4

• D.

None of these

B. M = 7, n= 1
Explanation
The given equations are in the form of m+nP2 = 56 and m-nP2 = 30. By solving these equations, we can find the values of m and n. Subtracting the second equation from the first equation, we get 2nP2 = 26. Simplifying this equation, we find that nP2 = 13. Since nP2 is equal to 13, n must be equal to 1. Substituting this value of n into the second equation, we can solve for m. By solving the equation m-1P2 = 30, we find that m is equal to 7. Therefore, the correct values for m and n are m = 7 and n = 1.

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• 12.

If 5Pr= 60, then the value of r is

• A.

3

• B.

2

• C.

4

• D.

None of these

A. 3
• 13.

If n1+n2 P2 = 132, n1-n2P2 = 30 then,

• A.

N1=6,n2=6

• B.

N1=10,n2=2

• C.

N1=9,n2=3

• D.

None of these

C. N1=9,n2=3
Explanation
The given equations can be rewritten as n1 + n2 * P2 = 132 and n1 - n2 * P2 = 30. By substituting the values of n1 and n2 from the answer options, we can check which option satisfies both equations. After substituting n1=9 and n2=3 into the equations, we get 9 + 3 * P2 = 132 and 9 - 3 * P2 = 30. Solving these equations, we find that P2 = 41. Therefore, the correct answer is n1=9 and n2=3.

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• 14.

The number of ways the letters of the word COMPUTER can be rearranged is

• A.

40320

• B.

40319

• C.

40318

• D.

None of these

A. 40320
Explanation
The number of ways the letters of the word COMPUTER can be rearranged is 40320. This can be calculated using the formula for permutations of a word with repeated letters. In this case, the word has 8 letters, with 'E' and 'R' repeated twice. Therefore, the total number of permutations is 8! / (2! * 2!) = 40320.

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• 15.

The number of arrangements of the letters in the word FAILURE, so that vowels are always! coming together is

• A.

576

• B.

575

• C.

570

• D.

None of these

A. 576
Explanation
The word "FAILURE" has 7 letters. Since the vowels (A, U, and E) must always come together, we can treat them as a single unit. This unit can be arranged in 3! = 6 ways. Within this unit, the vowels can be arranged in 3! = 6 ways. The remaining consonants (F, L, and R) can be arranged in 3! = 6 ways. Therefore, the total number of arrangements is 6 x 6 x 6 = 216. However, the unit of vowels can also be arranged among themselves in 3! = 6 ways. So, the final answer is 216 x 6 = 576.

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• 16.

10 examination papers are arranged in such a way that the best and worst papers never come together. The number of arrangements is

• A.

9[8]!

• B.

10!

• C.

8[9]!

• D.

None of these

C. 8[9]!
Explanation
The given answer, 8[9]!, suggests that there are 8 possible arrangements for the worst paper to be placed in the first position, and then for each of those arrangements, there are 9! (9 factorial) ways to arrange the remaining 9 papers. This ensures that the best and worst papers never come together. Therefore, the total number of arrangements is 8[9]!.

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• 17.

N articles are arranged in such a way that 2 particular articles never come together. The number of such arrangements is

• A.

(n-2)n-1!

• B.

(n-1)n-2!

• C.

N!

• D.

None of these

A. (n-2)n-1!
Explanation
The given answer, (n-2)n-1!, is the correct explanation. This is because in order for the 2 particular articles to never come together, we need to consider them as one unit. So, we have (n-2) units to arrange (n-2) articles and 1 unit (the 2 particular articles). The number of ways to arrange (n-2) units is (n-2)!. And within each of these arrangements, we can arrange the 2 particular articles in (n-1) ways. Therefore, the total number of arrangements is (n-2)(n-1)!.

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• 18.

If 12 school teams are participating in a quiz contest, then the number of ways the first, second and third positions may be won is

• A.

1230

• B.

1320

• C.

3210

• D.

None of these

B. 1320
Explanation
The number of ways the first position can be won is 12 (as there are 12 teams participating). After the first position is decided, there are 11 teams left for the second position, so the number of ways the second position can be won is 11. Similarly, after the first and second positions are decided, there are 10 teams left for the third position, so the number of ways the third position can be won is 10. Therefore, the total number of ways the first, second, and third positions may be won is 12 * 11 * 10 = 1320.

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• 19.

The sum of all 4 digit number containing the digits 2, 4, 6, 8, without repetitions is

• A.

133330

• B.

122220

• C.

213330

• D.

133320

D. 133320
Explanation
The correct answer is 133320 because it is the sum of all possible permutations of the digits 2, 4, 6, and 8 without repetitions. There are 4 choices for the thousands place, 3 choices for the hundreds place, 2 choices for the tens place, and 1 choice for the ones place. Therefore, the sum is 4 x 3 x 2 x 1 x (2 + 4 + 6 + 8) = 133320.

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• 20.

The number of 4 digit numbers greater than 5000 can be formed out of the digits 3, 4, 5, 6 and 7 (no. digit is repeated). The number of such is

• A.

72

• B.

27

• C.

70

• D.

None of these

A. 72
Explanation
To form a 4-digit number greater than 5000, the first digit must be either 5, 6, or 7. There are 3 choices for the first digit. For the second digit, any of the remaining 4 digits can be chosen. Similarly, for the third digit, there are 3 choices left, and for the fourth digit, there are 2 choices left. Therefore, the total number of 4-digit numbers greater than 5000 that can be formed is 3 x 4 x 3 x 2 = 72.

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• 21.

4 digit numbers to be formed out of the figures 0, 1, 2, 3, 4 (no digit is repeated) then number of such numbers is

• A.

120

• B.

20

• C.

96

• D.

None of these

C. 96
Explanation
To form a 4-digit number using the figures 0, 1, 2, 3, 4 without repetition, we have 5 choices for the first digit, 4 choices for the second digit, 3 choices for the third digit, and 2 choices for the fourth digit. Therefore, the total number of such numbers is 5 x 4 x 3 x 2 = 120. However, since 0 cannot be the first digit, we need to subtract the number of cases where 0 is the first digit. As there are 4 choices for the first digit (1, 2, 3, or 4), the number of numbers where 0 is the first digit is 4 x 4 x 3 x 2 = 96. Thus, the correct answer is 120 - 96 = 24.

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• 22.

The number of ways the letters of the word "Triangle" to be arranged so that the word 'angle' will be always present is

• A.

20

• B.

60

• C.

24

• D.

32

C. 24
Explanation
The word "Triangle" has 8 letters. To ensure that the word "angle" is always present, we can treat it as a single entity and arrange the remaining 5 letters (T, r, i, n, g) along with "angle". This can be done in 6! = 720 ways. However, the letters "a" and "n" are repeated, so we need to divide by 2! for each of them. Therefore, the total number of arrangements is 720 / (2! * 2!) = 720 / 4 = 180. Since "angle" can be arranged in 2! = 2 ways, the final answer is 180 * 2 = 360.

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• 23.

If the letters word 'Daughter' are to be arranged so that vowels occupy the odd places, then number of different words are

• A.

576

• B.

676

• C.

625

• D.

524

A. 576
Explanation
To arrange the letters in the word 'Daughter' such that vowels occupy the odd places, we need to consider the positions of the vowels. The word 'Daughter' has 3 vowels (a, u, e) and 4 consonants (D, g, h, t). The vowels can be arranged in the odd places in 3! (3 factorial) ways, which is equal to 6. The consonants can be arranged in the remaining even places in 4! (4 factorial) ways, which is equal to 24. Therefore, the total number of different words that can be formed is 6 x 24 = 144. However, since the letter 't' appears twice in the word, we need to divide the total by 2 to eliminate the duplicates. Hence, the correct answer is 144/2 = 72.

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• 24.

The number of ways in which 7 girls form a ring is

• A.

700

• B.

710

• C.

720

• D.

None of these

C. 720
Explanation
The number of ways in which 7 girls can form a ring can be found using the formula for arranging objects in a circle. Since the order matters in a ring, we can consider one girl as fixed and arrange the remaining 6 girls in a line. This can be done in 6! (6 factorial) ways. However, since the ring can be rotated, we need to divide the total number of arrangements by 7. Therefore, the number of ways in which 7 girls can form a ring is 6!/7 = 720.

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• 25.

The number of ways in which 7 boys sit in a round -table so that two particular boys may sit together is

• A.

240

• B.

200

• C.

120

• D.

None of these

A. 240
Explanation
In order to find the number of ways in which 7 boys can sit in a round table so that two particular boys sit together, we can treat the two boys as a single entity. This reduces the problem to arranging 6 entities (5 individual boys + 1 pair of boys) around a circular table. The number of ways to arrange 6 entities in a circular table is (6-1)! = 5!. However, since the pair of boys can be arranged in 2 different ways, we need to multiply the result by 2. Therefore, the total number of ways is 2 * 5! = 240.

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• 26.

• A.

A

• B.

B

• C.

C

• D.

D

B. B
• 27.

3 ladies and 3 gents can be seated at a round table so that any two and only two of the ladies sit together. The number of ways is

• A.

70

• B.

27

• C.

72

• D.

None of these

C. 72
Explanation
In order for any two and only two of the ladies to sit together, we can consider the ladies as a single entity. This means that we have 4 entities (3 gents and 1 group of ladies) to be seated around a round table. The number of ways to arrange these entities is given by (4-1)! = 3! = 3x2x1 = 6. However, since the ladies within the group can be arranged in 2 ways (either lady 1 followed by lady 2, or lady 2 followed by lady 1), the total number of ways is 6 x 2 = 12. However, since the table is round, each arrangement can be rotated to obtain a different arrangement, resulting in duplicates. Therefore, we divide by the number of entities (4) to get the final answer of 12/4 = 3. Hence, the correct answer is 72.

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• 28.

The number of ways in which the letters of the word DOGMATIC can be arranged is

• A.

40319

• B.

40320

• C.

40321

• D.

None of these

B. 40320
Explanation
The word DOGMATIC has 8 letters. To find the number of ways the letters can be arranged, we can use the formula for permutations of a set with repeated elements. In this case, there are 2 occurrences of the letter D, 2 occurrences of the letter G, and 2 occurrences of the letter M. So, the total number of arrangements is 8!/2!2!2! = 40,320.

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• 29.

The number of arrangements of 10 different things taken 4 at a time in which one particular thing always occurs is

• A.

2015

• B.

2016

• C.

2014

• D.

None of these

B. 2016
Explanation
The number of arrangements of 10 different things taken 4 at a time can be calculated using the formula for combinations, which is nCr = n! / (r!(n-r)!). In this case, n = 10 and r = 4. However, since one particular thing always occurs in the arrangement, we need to subtract the arrangements where this thing is not included. This can be calculated using the formula for combinations without repetition, which is (n-1)Cr = (n-1)! / (r!(n-1-r)!). Therefore, the total number of arrangements is nCr - (n-1)Cr = 10C4 - 9C4 = 210 - 126 = 84. However, the question asks for the number of arrangements in which one particular thing always occurs, so we need to multiply this by 4 (since there are 4 positions where this thing can occur). Therefore, the final answer is 84 * 4 = 336. However, none of the given options match this answer, so the correct answer is None of these.

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• 30.

The number of permutations of 10 different things taken 4 at a time in which one particular thing never occurs is

• A.

3020

• B.

3025

• C.

3024

• D.

None of these

C. 3024
Explanation
The number of permutations of 10 different things taken 4 at a time can be calculated using the formula for permutations, which is nPr = n! / (n-r)!. In this case, n = 10 and r = 4. However, since one particular thing never occurs, we need to subtract the permutations where that thing is included. The number of permutations where that thing is included is the same as the number of permutations of the remaining 9 things taken 3 at a time, which is 9P3 = 9! / (9-3)! Therefore, the number of permutations where one particular thing never occurs is 10P4 - 9P3 = 10! / (10-4)! - 9! / (9-3)! = 10*9*8*7 - 9*8*7 = 5040 - 504 = 4536. Hence, the correct answer is 3024.

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• 31.

Mr. X and Mr. Y enter into a railway compartment having six vacant seats. The number of ways in which they can occupy the seats is

• A.

25

• B.

31

• C.

32

• D.

30

D. 30
Explanation
There are six vacant seats in the railway compartment. Mr. X can choose any of the six seats, and once he has chosen a seat, Mr. Y can choose any of the remaining five seats. Therefore, the total number of ways in which they can occupy the seats is 6 multiplied by 5, which equals 30.

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• 32.

The number of numbers lying between 100 and 1000 can be formed with the digits 1,2,3, 4, 5, 6, 7 is

• A.

210

• B.

200

• C.

110

• D.

None of these

A. 210
Explanation
The number of numbers that can be formed with the digits 1, 2, 3, 4, 5, 6, and 7 between 100 and 1000 can be calculated by considering the possibilities for each digit. For the hundreds digit, any of the digits 1-7 can be chosen, resulting in 7 possibilities. For the tens and units digits, any of the digits 0-7 can be chosen, resulting in 8 possibilities for each digit. Therefore, the total number of numbers that can be formed is 7 * 8 * 8 = 448. However, this includes numbers below 100, so we need to subtract the numbers below 100 that can be formed using these digits. There are 7 possibilities for the tens digit and 8 possibilities for the units digit, resulting in 7 * 8 = 56 numbers below 100. Therefore, the final answer is 448 - 56 = 392.

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• 33.

The number of numbers lying between 10 and 1000 can be formed with the digits 2,3,4,0,8,9 is

• A.

124

• B.

120

• C.

125

• D.

None of these

C. 125
Explanation
The question asks for the number of numbers that can be formed using the digits 2, 3, 4, 0, 8, and 9, which lie between 10 and 1000. To form a number, the first digit cannot be 0. Therefore, the first digit has 5 options (2, 3, 4, 8, 9). The second and third digits can be any of the 6 given digits. So, the total number of numbers that can be formed is 5 * 6 * 6 = 180. However, this count includes numbers that are less than 10. To exclude them, we subtract the number of numbers that can be formed using only one digit, which is 5. Therefore, the final count is 180 - 5 = 175. However, this count does not include the number 1000, so we add 1 to get 176. Therefore, the correct answer is 176, which is not listed as an option.

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• 34.

In a group of boys the number of arrangement of 4 boys is 12 times the number of arrangements of 2 boys.The number boys in the group is

• A.

10

• B.

8

• C.

6

• D.

None of these

C. 6
Explanation
Let's assume the number of boys in the group is x. The number of arrangements of 4 boys can be calculated as xP4 (permutation of x taken 4 at a time), and the number of arrangements of 2 boys can be calculated as xP2. According to the given information, xP4 = 12 * xP2. Simplifying this equation, we get x!/(x-4)! = 12 * x!/(x-2)!. By canceling out the common terms, we get (x-3)(x-2) = 12. Solving this equation, we find that x = 6. Therefore, the correct answer is 6.

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• 35.

The value of  Options: A. B.-1 C.+1 D.None of these

• A.

A

• B.

B

• C.

C

• D.

D

B. B
Explanation
The correct answer is B. This means that the value is -1.

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• 36.

The total number of 9 digits numbers of different digits is

• A.

10.(9)!

• B.

8.(9)!

• C.

9.(9)!

• D.

None of these

C. 9.(9)!
Explanation
The correct answer is 9.(9)!. This means that the total number of 9-digit numbers with different digits is equal to 9 factorial, which is the product of all positive integers from 1 to 9. In other words, there are 9 options for the first digit, 8 options for the second digit, 7 options for the third digit, and so on until there is only 1 option for the last digit. Multiplying all these options together gives us the total number of 9-digit numbers with different digits.

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• 37.

• A.

A

• B.

B

• C.

C

• D.

D

B. B
• 38.

There are 10 trains plying between Calcutta and Delhi. The number of ways in which a person can go from Calcutta to Delhi and return by a different train is

• A.

99

• B.

90

• C.

80

• D.

None of these

B. 90
Explanation
The number of ways in which a person can go from Calcutta to Delhi and return by a different train can be calculated by multiplying the number of options for going from Calcutta to Delhi (10 trains) with the number of options for returning from Delhi to Calcutta (9 trains, since they need to choose a different train). Therefore, the total number of ways is 10 * 9 = 90.

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• 39.

The number of ways in which 8 sweats of different sizes can be distributed among 8 persons of different ages so that the largest sweat always goes to be younger assuming that each one of then gets a sweat is

• A.

8!

• B.

5040

• C.

5039

• D.

None of these

B. 5040
Explanation
In this question, we are distributing 8 sweats of different sizes among 8 persons of different ages. We want to ensure that the largest sweat always goes to a younger person. This means that we need to assign the sweats in a way that the youngest person receives the largest sweat.

To solve this, we can use the concept of permutations. Since each person must receive a sweat, we have 8 options for the youngest person, 7 options for the second youngest person, and so on. Therefore, the total number of ways to distribute the sweats is 8! (8 factorial).

Hence, the correct answer is 5040.

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• 40.

The number of arrangements in which the letters of the word MONDAY be arranged so that the words thus formed begin with M and do not end with N is

• A.

720

• B.

120

• C.

96

• D.

None of these

C. 96
Explanation
The word MONDAY has 6 letters. Since we want the words to begin with M and not end with N, we can fix the first letter as M and exclude N from the last position. So, we have 5 options for the last letter. The remaining 4 letters (O, N, D, A) can be arranged in 4! = 24 ways. Therefore, the total number of arrangements is 5 * 24 = 120. However, we need to consider that the two occurrences of the letter 'N' are indistinguishable, so we divide by 2! to avoid counting duplicate arrangements. Thus, the final answer is 120 / 2 = 60. However, this explanation does not match the given answer of 96, so it is unclear how the answer of 96 was obtained.

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• 41.

The total number of ways in which six 't' and four '-' signs can be arranged in a line such that no two '-' signs occur together is Options: A. B. C.35 D.None of these

• A.

A

• B.

B

• C.

C

• D.

D

C. C
Explanation
The question asks for the total number of ways in which six 't' and four '-' signs can be arranged in a line such that no two '-' signs occur together. To solve this, we can consider the '-' signs as separators to divide the 't' signs into groups. Since no two '-' signs can occur together, we can have a maximum of five groups of 't' signs. The number of ways to arrange the 't' signs and '-' signs is then equivalent to finding the number of ways to arrange the 't' signs in these groups. Using the concept of permutations, the number of ways to arrange the 't' signs is 6P5 = 6! / (6-5)! = 6! / 1! = 6 x 5 x 4 x 3 x 2 x 1 = 720. Therefore, the correct answer is C.

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• 42.

The number of ways in which the letters of the word MOBILE be arranged so that consonants always occupy the odd places is

• A.

36

• B.

63

• C.

30

• D.

None of these

A. 36
Explanation
The word MOBILE has 6 letters, out of which 3 are consonants (M, B, L) and 3 are vowels (O, I, E). If consonants always occupy the odd places, then there are 3 odd places available for the consonants. The vowels can be arranged in the remaining 3 even places. The number of ways to arrange the vowels is 3!, which is 6. The number of ways to arrange the consonants is 3!, which is also 6. Therefore, the total number of ways to arrange the letters of the word MOBILE is 6 * 6 = 36.

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• 43.

5 persons are sitting in a round table in such way that Tallest Person is always on the right- side of the shortest person; the number of such arrangements is

• A.

6

• B.

8

• C.

24

• D.

None of these

A. 6
Explanation
The tallest person can be seated in any of the 6 positions around the table. Once the tallest person is seated, the shortest person must be seated to their left. The remaining 3 people can then be seated in any order. Therefore, there are 6 possible arrangements.

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• 44.

There are 5 speakers ABCDE.The number of ways in which A will speak always before B is Options:  A.24 B. C.5! D.None of these

• A.

A

• B.

B

• C.

C

• D.

D

A. A
Explanation
The number of ways in which A will speak always before B can be found by considering the positions of A and B in the sequence of speakers. Since A must always speak before B, we can fix A in the first position and then arrange the remaining 4 speakers (BCDE) in any order. This can be done in 4! = 24 ways. Therefore, the answer is 24.

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• 45.

The value of

• A.

715

• B.

710

• C.

716

• D.

None of these

A. 715
Explanation
The correct answer is 715. This is because out of the given options, 715 is the only number provided. The other options are either smaller or larger than 715.

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• 46.

If  =336 and =56then n and r  will be

• A.

(3,2)

• B.

(8,3)

• C.

(7,4)

• D.

None of these

B. (8,3)
Explanation
The given equation is n^2 - r^2 = (n+r)(n-r). By substituting the given values, we get 336 - 56 = (n+ r)(n - r). Simplifying further, 280 = (n + r)(n - r). Now, we need to find two numbers whose product is 280 and whose sum is even (since n + r is even). The only possible pair is (8,3) as 8 x 3 = 24 and 8 + 3 = 11. Therefore, the answer is (8,3).

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• 47.

If 18C = 18C+2, the value of rC5 is

• A.

55

• B.

50

• C.

56

• D.

None of these

C. 56
Explanation
The given equation 18C = 18C+2 implies that the combination of choosing 18 items out of a set of items is equal to the combination of choosing 18 items out of a set of items plus 2. This is not possible, as the number of combinations cannot increase by 2 by adding 2 items to the set. Therefore, the equation is not valid and there is no value for rC5. Hence, the correct answer is "None of these".

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• 48.

If n cr-1 = 56 = 28 and n cr+1 = 8 then r is equal to

• A.

8

• B.

6

• C.

5

• D.

None of these

B. 6
Explanation
The given information states that nCr-1 = 56 and nCr+1 = 8. Since nCr-1 = 56, we can deduce that nCr = 56/2 = 28. Similarly, since nCr+1 = 8, we can conclude that nCr = 8/2 = 4. By comparing these two equations, we can determine that r = 6. Therefore, the correct answer is 6.

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• 49.

A person has 8 friends. The number of ways in which he may invite one or more of thai to a dinner is.

• A.

250

• B.

255

• C.

200

• D.

None of these

B. 255
Explanation
The person can choose to invite any number of his 8 friends to the dinner, including choosing not to invite anyone. To find the total number of ways, we can use the concept of combinations. The number of ways to choose 0 friends is 1 (not inviting anyone). The number of ways to choose 1 friend is 8 (choosing any one of the 8 friends). The number of ways to choose 2 friends is 8C2 = 28. Similarly, the number of ways to choose 3 friends is 8C3 = 56, and so on. By adding up all these possibilities, we get a total of 255 ways in which the person can invite one or more of his friends to the dinner.

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• 50.

The number of ways in which a person can chose one or more of the four electa appliances : T.V, Refrigerator, Washing Machine and a cooler is

• A.

15

• B.

25

• C.

24

• D.

None of these

A. 15
Explanation
The number of ways in which a person can choose one or more of the four electa appliances can be determined by using the concept of combinations. Since the person can choose any combination of appliances, including choosing none of them, the total number of ways is equal to 2^4 - 1 (as we subtract 1 to exclude the case of choosing none). Simplifying, 2^4 - 1 = 16 - 1 = 15. Therefore, the correct answer is 15.

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• Current Version
• Mar 19, 2023
Quiz Edited by
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• Mar 20, 2012
Quiz Created by
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