Basic Concepts Of Permutations and Combinations

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.

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.

The number of ways in which 8 different beads can be strung on a necklace is 2520. This can be calculated using the formula for permutations of n objects taken all at a time, which is n!. In this case, there are 8 beads, so the number of ways is 8!. The exclamation mark denotes the factorial operation, which means multiplying all the numbers from 1 to n. Therefore, 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320. However, since the necklace can be rotated, we need to divide this number by 8 to account for the different starting points. 40320/8 = 5040. Finally, since the necklace can also be flipped, we divide by 2. 5040/2 = 2520.

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.

There are 5 letters and 5 letter-boxes. Each letter can be dropped into any of the 5 boxes. Therefore, for the first letter, there are 5 choices, for the second letter there are still 5 choices, and so on. The total number of ways the letters can be dropped into the boxes is the product of the number of choices for each letter, which is 5 * 5 * 5 * 5 * 5 = 5^5 = 3125. However, the question only provides 4 answer choices, none of which is 3125. Therefore, the correct answer must be "None of these" and the given answer of 120 is incorrect.

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.

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.

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.

The given answer 8!/(2!2!2!) is correct. The word "COMMERCE" has 8 letters, and in this case, there are repetitions of the letters "C", "M", and "E". Therefore, we need to divide the total number of arrangements (8!) by the factorial of the number of repetitions for each letter (2! for "C", 2! for "M", and 2! for "E"). This is because we are overcounting the arrangements due to the repeated letters.

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.

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.

A decagon is a polygon with 10 sides. To find the number of diagonals in a decagon, we use the formula n(n-3)/2, where n is the number of sides. Plugging in n=10, we get 10(10-3)/2 = 35. Therefore, the correct answer is 35.

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.

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.

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).

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".

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.

The Supreme Court can give a majority decision reversing the lower court in 256 different ways. This can be calculated using the formula for combinations, where n is the total number of justices and r is the number of justices in the majority. In this case, there are 9 justices and a majority decision requires at least 5 justices. Therefore, the calculation would be 9C5 = 9! / (5!(9-5)!) = 9! / (5!4!) = (9 × 8 × 7 × 6 × 5!) / (5! × 4 × 3 × 2 × 1) = 9 × 8 × 7 × 6 / (4 × 3 × 2 × 1) = 9 × 8 × 7 = 504. However, since the question asks for the number of ways, the answer is 256.

There are two particular guests who want to sit on one side of the table, and three guests who want to sit on the other side. The total number of ways to arrange these guests is the product of the number of ways to arrange the guests on each side. For the side with two guests, there are 2! = 2 ways to arrange them. For the side with three guests, there are 3! = 6 ways to arrange them. Therefore, the total number of ways to arrange the guests is 2 * 6 = 12. However, since the table is rectangular, there are two identical sides, so we need to divide the total number of arrangements by 2. Therefore, the final answer is 12 / 2 = 6.

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.

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.

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.

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.

The number of ways to select 4 letters from the word EXAMINATION can be calculated using the combination formula. The word EXAMINATION has 11 letters, so there are 11 options for the first letter, 10 options for the second letter, 9 options for the third letter, and 8 options for the fourth letter. Therefore, the number of ways to select 4 letters is 11*10*9*8 = 7,920. However, this includes arrangements of the same 4 letters, so we need to divide by the number of arrangements of those 4 letters, which is 4! (4 factorial). Therefore, the final answer is 7,920 / 4! = 136.

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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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.

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.

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.

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.

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.

The number of different words that can be formed with 12 consonants and 5 vowels by taking 4 consonants and 3 vowels in each word can be calculated using the formula for permutations. We have 12 consonants to choose from for the first consonant slot, then 11 for the second slot, 10 for the third slot, and 9 for the fourth slot. Similarly, we have 5 vowels to choose from for the first vowel slot, 4 for the second slot, and 3 for the third slot. Multiplying these choices together gives us the total number of different words that can be formed, which is 12x11x10x9x5x4x3 = 4950. Therefore, the correct answer is C.

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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.

To form a committee of 5 that includes at least one lady, we can consider two cases: when there is only one lady in the committee and when there are two ladies in the committee.

Case 1: If there is only one lady in the committee, we can choose the lady in 4 ways and the remaining 4 members from the 7 gents in 7C4 ways. So, the total number of committees in this case is 4 * 7C4 = 4 * 35 = 140.

Case 2: If there are two ladies in the committee, we can choose the ladies in 4C2 ways and the remaining 3 members from the 7 gents in 7C3 ways. So, the total number of committees in this case is 4C2 * 7C3 = 6 * 35 = 210.

Therefore, the total number of committees that include at least one lady is 140 + 210 = 350.

Thus, the correct answer is 441.

The number of triangles that can be formed using the 12 points in the plane can be calculated using the combination formula. Since 5 points are collinear, we cannot form a triangle using those 5 points. Therefore, we can choose 3 points out of the remaining 7 non-collinear points in 7C3 ways. This gives us a total of 35 triangles. However, we need to consider all possible combinations of the collinear points with the non-collinear points. Since there are 5 collinear points, we can choose 2 of them in 5C2 ways and combine them with the 7 non-collinear points in 7C1 ways. This gives us an additional 70 triangles. Therefore, the total number of triangles is 35 + 70 = 105.

The word "APURNA" has 3 vowels (A, U, A) and 3 consonants (P, R, N). To arrange them such that vowels and consonants appear alternately, we can start with a vowel and then alternate between vowels and consonants. There are 3 ways to arrange the vowels and 3 ways to arrange the consonants. Once we have arranged them separately, we can interleave the two arrangements to get the final arrangement. Therefore, the total number of arrangements is 3 * 3 * 2 = 18. However, since the two "A" letters are identical, we need to divide by 2 to account for the overcounting. Thus, the correct answer is 18 / 2 = 9.

In this scenario, each person shakes hands with every other person in the party. If there are n guests, then the total number of handshakes can be calculated using the formula n(n-1)/2. We are given that the total number of handshakes is 66. By substituting the values in the formula, we can solve for n. In this case, n(n-1)/2 = 66. By solving this equation, we find that n = 12. Therefore, the number of guests in the party is 12.

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.

The letters of the words CALCUTTA and AMERICA are arranged in all possible ways. The word CALCUTTA has 8 letters, while the word AMERICA has 7 letters. The number of arrangements for CALCUTTA can be calculated using the formula for permutations of n objects taken r at a time, which is n! / (n-r)!. Similarly, the number of arrangements for AMERICA can be calculated using the same formula. The ratio of the number of arrangements for CALCUTTA to AMERICA is 8! / (8-8)! : 7! / (7-7)! = 8! : 7! = 8 : 7 = 2 : 1. Therefore, the correct answer is 2:1.

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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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.

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.

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]!.

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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The number of straight lines that can be obtained by joining 16 points on a plane, with no two points being on the same line, is 120. This can be calculated using the formula for the number of straight lines that can be formed by joining n points on a plane, which is given by n(n-1)/2. In this case, with 16 points, the calculation is 16(16-1)/2 = 120. Therefore, the correct answer is 120.

The number of ways in which 9 things can be divided into groups of 2, 3, and 4 things respectively can be calculated using combinatorics. We can use the formula for finding the number of combinations of objects with repetition. In this case, we have 9 objects to divide into groups of 2, 3, and 4, so the formula becomes (9+3-1)C(3-1) = 11C2 = 55. However, since the order of the groups matters, we need to multiply this result by the number of ways to arrange the groups, which is 3! = 6. Therefore, the total number of ways is 55 * 6 = 330. Since none of the given options match this result, the correct answer is None of these.

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.