Question 1

A committee of 5 members is to be formed out of 3 trainees, 4 professors and 6 research associates. In how many different ways can this be done, if the committee should have 4 professors and 1 research associate or all 3 trainees and 2 professors?

Accepted Answer

Concept & Addition Principle: Use combinations ^nC_r to select group members. The word "and" indicates multiplication between selections, while "or" indicates addition between separate, mutually exclusive cases. Step-by-Step Solution: 1. Case 1: 4 professors AND 1 research associate. Select 4 professors out of 4: ^4C_4 = 1 way. Select 1 research associate out of 6: ^6C_1 = 6 ways. Ways for Case 1 = 1 × 6 = 6. 2. Case 2: 3 trainees AND 2 professors. Select 3 trainees out of 3: ^3C_3 = 1 way. Select 2 professors out of 4: ^4C_2 = (4 × 3/2 × 1) = 6 ways. Ways for Case 2 = 1 × 6 = 6. 3. Total Ways (Case 1 OR Case 2): Total = 6 + 6 = 12 ways. Exam Strategy & Shortcut: Recognize that selecting all members of a specific subgroup (like 4 professors out of 4) is always just 1 way (^nC_n = 1). This immediately simplifies the multiplication to 1 × 6 for both cases. Common Pitfall: Multiplying the results of Case 1 and Case 2 instead of adding them. Remember: OR means ADD. Final Answer: Therefore, the correct answer is 12.

Question 2

In how many ways can a committee of 4 people be chosen out of 8 people?

Accepted Answer

Concept & Combination Basics: Choosing a subgroup from a larger group without respect to arrangement order is a classic combination problem, solved using: ^nC_r = (n!/r!(n-r)!) Step-by-Step Solution: 1. Identify total people available: n = 8. 2. Identify the number of people to choose: r = 4. 3. Apply the combination formula: ^8C_4 = (8 × 7 × 6 × 5/4 × 3 × 2 × 1) 4. Simplify the expression: 4 × 2 in the denominator cancels the 8 in the numerator, and 3 cancels into 6 (leaving 2). 7 × 2 × 5 = 70 Exam Strategy & Shortcut: To calculate ^nC_r quickly, expand the numerator with r descending terms and the denominator as r!. For ^8C_4, just write out (8 × 7 × 6 × 5/24) and reduce it mentally. Common Pitfall: Using permutation (^8P_4 = 1680) instead of combination. Committees do not imply any specific roles or order, so always use combinations. Final Answer: Therefore, the correct answer is 70.

Question 3

A select group of 4 is to be formed from 8 men and 6 women in such a way that the group must have at least 1 woman. In how many different ways can it be done?

Accepted Answer

Concept & Complementary Counting: When tasked with finding combinations with "at least 1" item from a subgroup, calculate the total unrestricted combinations and subtract the combinations where the subgroup is completely excluded. Step-by-Step Solution: 1. Total Unrestricted Groups: We are selecting 4 people from a total pool of 14 people (8 men + 6 women). ^{14}C_4 = (14 × 13 × 12 × 11/4 × 3 × 2 × 1) = 1001 ways 2. Groups with NO Women: Select 4 people entirely from the 8 men. ^8C_4 = (8 × 7 × 6 × 5/4 × 3 × 2 × 1) = 70 ways 3. Groups with At Least 1 Woman: Subtract the "no women" scenario from the total. 1001 - 70 = 931 ways Exam Strategy & Shortcut: Notice the denominators when expanding combinations: 4 × 3 × 2 × 1 = 24. In ^{14}C_4, 12 in the numerator cancels beautifully with 24 to leave 2, which further cancels with 14 to leave 7. Factoring out rather than multiplying large numbers prevents arithmetic mistakes. Common Pitfall: Failing to combine the total pool correctly (8+6=14) or incorrectly applying permutation formula instead of combination. Final Answer: Therefore, the correct answer is 931.

Question 4

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

Accepted Answer

Concept & Permutations with Repetition: When arranging a set of items where some items are identical, use the formula for permutations with indistinguishable objects: (n!/p_1! × p_2! × \dots × p_k!) where n is the total number of items, and p_1, p_2 are the frequencies of repeating items. Step-by-Step Solution: 1. Count the total number of letters in "ENGINEERING": n = 11. 2. Identify and count the repeating letters: - E appears 3 times. - N appears 3 times. - G appears 2 times. - I appears 2 times. - R appears 1 time (does not affect division). 3. Apply the formula: (11!/3! × 3! × 2! × 2!) 4. Expand and simplify: (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1/6 × 6 × 2 × 2) Denominator equals 144. 11! = 39,916,800. (39916800/144) = 277200 Exam Strategy & Shortcut: To simplify 11! / 144 manually, do not multiply out 11!. Cancel factors! The 6 × 6 in the denominator (3! × 3!) cancels 36 from the numerator (e.g., 9 × 4). The 2 × 2 = 4 cancels into the 8 to leave 2. Multiply the remaining smaller integers directly. Common Pitfall: Missing a repeated letter. It is highly recommended to cross out letters as you tally their frequencies to ensure p_1 + p_2 + \dots = n. Final Answer: Therefore, the correct answer is 277200.

Question 5

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

Accepted Answer

Concept & Permutations with Identical Items: When calculating arrangements of words with repeated letters, divide the total factorial of the letter count by the factorials of the frequencies of the repeating letters. Arrangements = (n!/p_1! × p_2! \dots) Step-by-Step Solution: 1. Count the total letters in "ALLAHABAD": n = 9. 2. Tally the frequencies of the letters: - A: 4 - L: 2 - H: 1 - B: 1 - D: 1 3. Substitute these values into the permutation formula: (9!/4! × 2!) 4. Expand and cancel terms: (9 × 8 × 7 × 6 × 5 × 4!/4! × 2) = (9 × 8 × 7 × 6 × 5/2) 5. Calculate the remaining product: 9 × 4 × 7 × 6 × 5 = 36 × 210 = 7560 Exam Strategy & Shortcut: Stop expanding the numerator as soon as it matches the largest factorial in the denominator. Here, stop at 4! so it immediately cancels out with the 4! below, saving massive amounts of computation time. Common Pitfall: A frequent error is dividing by just the number of identical letters (e.g., dividing by 4 and 2) rather than their factorials (4! and 2!). Final Answer: Therefore, the correct answer is 7560.

Question 6

In how many different ways can the letters of the word JUDGE be arranged in such a way that the vowels always come together?

Accepted Answer

Concept & Permutations with Grouping: When a specific set of items must always appear together, we treat them as a single unit or "block". We then find the permutations of all units, and multiply this by the permutations of the items within the "block". The general formula for arranging n distinct items is n!. If items are grouped into 1 unit, the total units become (n - k + 1) where k is the size of the group. Step-by-Step Solution: • Analyze the Word: The word "JUDGE" contains 5 letters. • Identify Vowels and Consonants: • Vowels: U, E (2 letters) • Consonants: J, D, G (3 letters) • Group the Vowels: Since the vowels must be together, treat (UE) as a single unit. • Count the Units: We now have the consonants J, D, G and the vowel unit (UE). This gives a total of 3 + 1 = 4 units. • Arrange the Units: The 4 units can be arranged in 4! ways. • 4! = 4 × 3 × 2 × 1 = 24 • Arrange Items Within the Unit: The 2 vowels (U, E) can be arranged among themselves in 2! ways. • 2! = 2 × 1 = 2 • Calculate Total Arrangements: Multiply the two results. • Total ways = 24 × 2 = 48 Exam Strategy & Shortcut: For simple non-repeating words, immediately write down (Total Consonants + 1)! × (Total Vowels)!. Here: 4! × 2! = 24 × 2 = 48. This takes less than 15 seconds. Common Pitfall: A common mistake is forgetting to arrange the vowels within their own group. Students often calculate 4! = 24 and stop there, resulting in an incorrect answer. Always account for internal permutations of the grouped block. Final Answer: Therefore, the correct answer is 48.

Question 7

In how many different ways can the letters of the word AUCTION be arranged in such a way that the vowels always come together?

Accepted Answer

Concept & Permutations with Grouping: To ensure specific items are always adjacent, group them into a single super-entity. The total arrangements equal the permutations of all entities multiplied by the internal permutations of the super-entity. Step-by-Step Solution: • Analyze the Word: The word "AUCTION" contains 7 letters. • Identify Vowels and Consonants: • Vowels: A, U, I, O (4 letters) • Consonants: C, T, N (3 letters) • Group the Vowels: Treat the 4 vowels (A, U, I, O) as a single unit. • Count the Units: We have the consonants C, T, N and the vowel unit (AUIO), making 3 + 1 = 4 units in total. • Arrange the Units: The 4 units can be arranged in 4! ways. • 4! = 24 • Arrange Items Within the Unit: The 4 vowels can be arranged internally in 4! ways. • 4! = 24 • Calculate Total Arrangements: Multiply the independent arrangements. • Total ways = 24 × 24 = 576 Exam Strategy & Shortcut: Recognize that neither the vowels nor the consonants have repeating letters. Apply the direct logical formula: (C+1)! × V!, where C = 3 and V = 4. So, 4! × 4! = 576. Common Pitfall: Mistaking the vowel count. Missing one vowel while scanning the word (like missing the 'I' or 'O') changes the calculation entirely. Always cross-check your letter tally with the original word length. Final Answer: Therefore, the correct answer is 576.

Question 8

In how many different ways can the letters of the word SOFTWARE be arranged in such a way that the vowels always come together?

Accepted Answer

Concept & Permutations with Grouping: When grouping elements, the grouped elements are treated as a single structural unit. The fundamental principle of counting implies we multiply the permutations of the macro-structure by the permutations of the micro-structure (the grouped elements). Step-by-Step Solution: • Analyze the Word: The word "SOFTWARE" contains 8 letters. • Identify Vowels and Consonants: • Vowels: O, A, E (3 letters) • Consonants: S, F, T, W, R (5 letters) • Group the Vowels: Treat the 3 vowels (O, A, E) as a single unit. • Count the Units: We have 5 consonants + 1 vowel unit = 6 total units. • Arrange the Units: The 6 units can be arranged in 6! ways. • 6! = 720 • Arrange Items Within the Unit: The 3 vowels can be internally arranged in 3! ways. • 3! = 6 • Calculate Total Arrangements: • Total ways = 720 × 6 = 4320 Exam Strategy & Shortcut: Memorize basic factorials up to 7! to save time. Knowing 6! = 720 and 3! = 6 lets you immediately calculate 720 × 6 = 4320 without writing out expanded multiplication steps. Common Pitfall: A common error is grouping the vowels but counting the remaining consonants incorrectly. Always double-check that your total letter count matches the sum of vowels and consonants. Final Answer: Therefore, the correct answer is 4320.

Question 9

In how many different ways can the letters of the word BANKING be arranged in such a way that the vowels always come together?

Accepted Answer

Concept & Permutations with Repetition: When arranging a set of items where some items are identical, the standard factorial is divided by the factorials of the counts of each identical item to remove duplicate permutations. Formula for arrangements of n items with p identical items of one kind: (n!/p!) Step-by-Step Solution: • Analyze the Word: The word "BANKING" contains 7 letters. • Identify Vowels and Consonants: • Vowels: A, I (2 letters) • Consonants: B, N, K, N, G (5 letters) • Notice that the letter 'N' repeats 2 times. • Group the Vowels: Treat (A, I) as 1 unit. • Count the Units: 5 consonants + 1 vowel unit = 6 units. • Arrange the Units: Since 'N' is repeated twice among the consonants, the arrangement of these 6 units requires dividing by 2!. • Arrangements of units = (6!/2!) = (720/2) = 360 • Arrange Items Within the Unit: The 2 vowels (A, I) arrange in 2! ways. • 2! = 2 • Calculate Total Arrangements: • Total ways = 360 × 2 = 720 Exam Strategy & Shortcut: Whenever there are repeating letters, handle the repetitions immediately in the respective block (either the main unit arrangement or the internal vowel arrangement). Don't wait until the final multiplication step to divide, as numbers get unnecessarily large. Common Pitfall: A major trap is forgetting to divide by the factorial of repeating letters, which would lead to an inflated answer (in this case, 1440). Always scan the word for duplicate letters before calculating. Final Answer: Therefore, the correct answer is 720.

Question 10

In how many different ways can the letters of the word MACHINE be arranged so that the vowels may occupy only the odd positions?

Accepted Answer

Concept & Positional Permutations: When specific items are restricted to specific positions, calculate the permutations for those constrained items first, then arrange the remaining items in the leftover positions. Formula for arranging r items in n available slots: ^nP_r = (n!/(n-r)!) Step-by-Step Solution: • Analyze the Word: "MACHINE" contains 7 letters. • Identify Vowels and Consonants: • Vowels: A, I, E (3 letters) • Consonants: M, C, H, N (4 letters) • Identify Odd Positions: In a 7-letter word, the odd positions are 1st, 3rd, 5th, and 7th. There are 4 odd positions. • Arrange Vowels: The 3 vowels must be placed in the 4 available odd positions. • Number of ways = ^4P_3 = 4 × 3 × 2 = 24 ways. • Arrange Consonants: After placing the 3 vowels, there is 1 odd position left, plus the 3 even positions (2nd, 4th, 6th). This gives a total of 4 remaining positions for the 4 consonants. • Number of ways = 4! = 24 ways. • Calculate Total Arrangements: Multiply the independent arrangements. • Total ways = 24 × 24 = 576. Exam Strategy & Shortcut: Deconstruct the problem into independent tasks: 3 vowels into 4 slots is 4 × 3 × 2 = 24. The remaining 4 letters into the remaining 4 slots is 4! = 24. Mentally calculate 24^2 = 576. Memorizing squares up to 25 saves significant time here. Common Pitfall: A frequent error is assuming the consonants must only go into the even positions. The problem states vowels must occupy *only* odd positions, but it doesn't restrict consonants from occupying the leftover odd position. Final Answer: Therefore, the correct answer is 576.

Question 11

There are six teachers. Out of them two are primary teachers and two are secondary teachers. They are to stand in a row, so as the primary teachers, middle teachers and secondary teachers are always in a set. The number of ways in which they can do …

Accepted Answer

Concept & Permutations of Multiple Groups: When multiple distinct groups must remain together, treat each entire group as a single super-unit. Calculate the permutations of the super-units, and multiply by the internal permutations within every individual super-unit. Step-by-Step Solution: • Identify the Sets of Teachers: • Total teachers = 6. • Primary teachers = 2. • Secondary teachers = 2. • Middle teachers = Total - Primary - Secondary = 6 - 2 - 2 = 2. • Form the Groups: • Group 1: Primary teachers (2 people) • Group 2: Middle teachers (2 people) • Group 3: Secondary teachers (2 people) • Arrange the Groups: • There are 3 distinct groups. They can be arranged in a row in 3! ways. • 3! = 6 ways. • Arrange Teachers Internally Within Groups: • Primary teachers can swap places: 2! = 2 ways. • Middle teachers can swap places: 2! = 2 ways. • Secondary teachers can swap places: 2! = 2 ways. • Calculate Total Arrangements: • Total ways = 3! × 2! × 2! × 2! • Total ways = 6 × 2 × 2 × 2 = 48. Exam Strategy & Shortcut: Recognize the macro/micro structure instantly. 3 blocks \rightarrow 3! = 6. Each block has 2 items \rightarrow 2^3 = 8. Total = 6 × 8 = 48. Common Pitfall: A common mistake is forgetting to deduce the number of middle teachers. While it isn't explicitly stated as "two", basic arithmetic (6 - 2 - 2) is required to realize there are 2 middle teachers making up the final set. Final Answer: Therefore, the correct answer is 48.

Question 12

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

Accepted Answer

Concept & Basic Permutation: The number of ways to arrange n distinct objects in a sequence is simply n factorial, denoted as n!. n! = n × (n-1) × (n-2) × \dots × 1 Step-by-Step Solution: • Analyze the Word: The word "BAKERY" consists of the letters B, A, K, E, R, Y. • Count the Letters: There are 6 letters in total. • Check for Repetitions: All letters are completely distinct (no letter appears more than once). • Calculate Arrangements: Since there are 6 distinct letters, the total number of permutations is 6!. • 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720. Exam Strategy & Shortcut: Factorial memorization is non-negotiable for competitive exams. You should instantly know that 5! = 120, 6! = 720, and 7! = 5040. This turns a calculation problem into a simple recall problem, taking less than 2 seconds. Common Pitfall: Checking for repeating letters is a mandatory first step. If a student assumes all letters are distinct without checking and uses n!, they will get the wrong answer on words like "BANANA". Fortunately, BAKERY has no repeats, so the basic formula applies safely. Final Answer: Therefore, the correct answer is 720.

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