Question 1

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 2

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.

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