Permutations & CombinationshardFree

Permutations & Combinations: Let Number Words Can Formed Using Letters Word

JEE Maths question with a full step-by-step solution.

Question

Let

N

be the number of words which can be formed using all the letters of the word "DARJEELING" so that there are at least two consonants between any two vowels. Then match List I with List II:

List-I

I If

N

is divisible by

2^{n}

\left(n \in \mathbb{N}\right)

, then

n

can be

IIIf

N

is divisible by

6^{P}

\left(P \in \mathbb{N}\right)

, then

P

must be less than

III Number of odd divisors of

N

is greater than

IV Number of zeroes at the end of

N

is less than

List-II

1

2

3

4

5

\to

(P, Q, R); II

\to

(Q, R, S, T); III

\to

(P, Q, R, S, T); IV

\to

(P, S, T)

\to

(Q, R, T); II

\to

(P, Q, T); III

\to

(P, Q, R, S); IV

\to

(P, Q, R, S, T)

\to

(P, Q, R, S, T); II

\to

(S, T); III

\to

(P, Q, R, S, T); IV

\to

(Q, R, S, T)correct

\to

(P, Q, S, T); II

\to

(P, Q, R, S, T); III

\to

(P, Q, R, T); IV

\to

(P, Q, R)

Solution

Step 1:

\text{Vowels: } A,\,E,\,E,\,I \ \left(4,\ \text{with } E \text{ twice}\right),\qquad \text{Consonants: } D,\,R,\,J,\,L,\,N,\,G \ \left(6,\ \text{all distinct}\right)

Step 2: Arrange the

6

consonants in a row; they create

7

gaps (before, between, after). Two vowels in the same gap would be adjacent with no consonant between them, so each of the four vowels occupies a different gap. Label the chosen gaps

0 \le g_1<g_2<g_3<g_4 \le 6

; then the number of consonants between the vowels in gaps

g_i

and

g_{i+1}

is exactly

g_{i+1}-g_i

. Step 3:

g_{i+1}-g_i \ \ge\ 2 \quad\text{for } i = 1,2,3

g_4 \ge g_1+6

, and since

0 \le g_1

and

g_4 \le 6

, the only possibility is

g_1 = 0,\quad g_2 = 2,\quad g_3 = 4,\quad g_4 = 6

i.e. the single pattern

V\,C\,C\,V\,C\,C\,V\,C\,C\,V

. Step 4:

N = \underbrace{\frac{4!}{2!}}_{\text{vowels, } E \text{ repeated}}\times\underbrace{6!}_{\text{consonants}} = 12\times720 = 8640

8640 = 2^{6}\cdot3^{3}\cdot5

(Exhaustive enumeration of all

\frac{10!}{2!} = 1814400

distinct arrangements of DARJEELING confirms

N = 8640

.) Step 5: Part (I),

N

divisible by

2^{n}

. The exponent of

2

6

, so

n

may be any of

1,2,3,4,5,6

. Every value in List-II qualifies:

\textbf{(I)} \to \textbf{P, Q, R, S, T}

Step 6: Part (II),

N

divisible by

6^{P} = 2^{P}3^{P}

. This needs

P \le 6

(from the

2

s) and

P \le 3

(from the

3

s), so

P \le 3

. The listed values that

P

**must** be less than are those exceeding

3

\textbf{(II)} \to \textbf{S, T}

(

P<3

fails because

P=3

is allowed;

P<1

and

P<2

fail likewise.) Step 7: Part (III), number of odd divisors. The odd part of

N

3^{3}\cdot5

, with

\left(3+1\right)\left(1+1\right) = 8 \ \text{divisors}

and

8

is greater than every value in List-II:

\textbf{(III)} \to \textbf{P, Q, R, S, T}

Step 8: Part (IV), number of trailing zeroes of

N

N = 8640

ends in exactly one zero (equivalently

\min

of the exponents of

2

and

5

\min(6,1) = 1

). The listed values it is **less than** are

2,3,4,5

; the bound

1

is excluded, since

1<1

is false:

\textbf{(IV)} \to \textbf{Q, R, S, T}

Step 9:

\text{I}\to\text{PQRST};\quad \text{II}\to\text{ST};\quad \text{III}\to\text{PQRST};\quad \text{IV}\to\text{QRST}

which is code (3). Answer: (3)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.