Permutations & Combinations Question - None

Question

Let N be the number of words which can be formed using all the letters of the word 'DARJEELING' so that there are atleast two consonants between any two vowels.

Match List-I with List-II:

List-I	List-II
(P) If N is divisible by $2^n$ ($n \in N$), then n can be	(1) 1
(Q) If N is divisible by $6^p$ ($p \in N$), then p must be less than	(2) 2
(R) Number of odd divisors of N is greater than	(3) 3
(S) Number of zeroes at the end of N is less than	(4) 4
	(5) 5
	(6) 6

Accepted Answer

Step 1: Identify letters in 'DARJEELING'

Vowels: A, E, E, I (4 vowels, with E repeated twice)

Consonants: D, R, J, L, N, G (6 consonants, all distinct)

Step 2: Apply the constraint (≥2 consonants between any two vowels)

With 4 vowels and the requirement of at least 2 consonants between consecutive vowels, we need to place vowels in the arrangement with gaps.

The pattern is: C...C V C C V C C V C C V C...C (where V = vowel, C = consonant)

We have 6 consonants to distribute in 5 positions (before first vowel, between each pair of vowels, after last vowel) with at least 2 in the 4 middle positions.

Step 3: Count valid arrangements

First, arrange 6 distinct consonants: 6! ways

Next, place 4 vowels in the 5 available gaps created by consonants, choosing 4 of 5 gaps with ≥2 consonants between any two selected vowels.

The valid placements require picking 4 positions from 5 such that no two are adjacent (which automatically ensures ≥2 consonants between vowels).

This gives C(5,4) = 5 ways to position the vowels.

Then arrange the vowels: 4!/2! = 12 ways (accounting for repeated E)

Step 4: Calculate N

N = 6! × 5 × (4!/2!) = 720 × 5 × 12 = 43,200

Step 5: Prime factorization of N

N = 43,200 = 432 × 100 = 16 × 27 × 100 = 2^4 × 3^3 × 2^2 × 5^2 = 2^6 × 3^3 × 5^2

Step 6: Analyze each statement

(P) N divisible by 2^n: Since N = 2^6 × 3^3 × 5^2, we have n ≤ 6. So n can be 1, 2, 3, 4, 5, 6. → P → {1,2,3,4,5,6}

(Q) N divisible by 6^p: 6^p = 2^p × 3^p. Since N has 2^6 × 3^3, we need p ≤ min(6,3) = 3, so p ≤ 3. Thus p must be less than 4. → Q → {4,5,6}

(R) Number of odd divisors: Odd divisors come from 3^3 × 5^2 only. Count = (3+1)(2+1) = 12. We need odd divisors > k. Since we have 12 odd divisors, the statement is true for k = 1,2,3,4,5,6,7,8,9,10,11. Matching with options: > 2, >3, >4, >5, >6. → R → {2,3,4,5,6}

(S) Trailing zeros: Zeros come from factors of 10 = 2 × 5. We have 2^6 × 5^2, so min(6,2) = 2 zeros. Number of zeros < k means k > 2. → S → {2,3,4,5,6}

∴ Answer: D

Question:

Step-by-Step Solution

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