The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy the even

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Concept understanding — Permutations Without Repetition

Permutations Without Repetition – The Idea of Arranging Things

Imagine you have three different books on a shelf: a Physics book, a Chemistry book, and a Maths book. How many different ways can you arrange them in a row?

You could try listing them out:

Physics, Chemistry, Maths
Physics, Maths, Chemistry
Chemistry, Physics, Maths
Chemistry, Maths, Physics
Maths, Physics, Chemistry
Maths, Chemistry, Physics

That's 6 arrangements. Notice that each arrangement uses all three books exactly once — no book is repeated, and no book is left out. This is the core idea: permutations without repetition count the number of ways to arrange a set of distinct objects in order, using each object exactly once.

Why "Without Repetition"?

The phrase "without repetition" means that once you place an object in a position, you cannot use it again. In our book example, once you put the Physics book in the first slot, you cannot put it in the second or third slot. Each object appears exactly once in the arrangement.

This is different from "permutations with repetition" (like creating 3-letter codes from the letters A, B, C where you can reuse letters — e.g., AAA, AAB, etc.). Here, no repeats allowed.

The Counting Logic – Why Multiply?

Let's build the arrangement step by step for 3 distinct books:

First position: You have 3 choices (any of the 3 books).
Second position: After placing the first book, only 2 books remain — so 2 choices.
Third position: Only 1 book is left — so 1 choice.

Total arrangements = $3 \times 2 \times 1 = 6$ .

This product $3 \times 2 \times 1$ is called 3 factorial, written as $3!$ .

P(n) = n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

For $n$ distinct objects, the number of permutations (arrangements in order) is $n!$ .

What If You Only Arrange Some of Them?

Suppose you have 5 different books, but you only want to arrange 3 of them on a shelf. How many ways?

First position: 5 choices
Second position: 4 choices
Third position: 3 choices

Total = $5 \times 4 \times 3 = 60$ .

This is a permutation of 5 objects taken 3 at a time, written as $P(5,3)$ or ${}^5P_3$ .

P(n, r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (n-r+1)

Here $n$ is the total number of distinct objects, and $r$ is how many you are arranging. The formula works because:

Numerator $n!$ counts all arrangements of all $n$ objects.
Denominator $(n-r)!$ removes the arrangements of the $n-r$ objects you are not using.

Key Points to Remember

Important

Order matters — swapping two objects gives a different permutation.
No repetition — each object is used at most once.
For arranging all $n$ objects: $n!$
For arranging r out of $n$ objects: $\frac{n!}{(n-r)!}$

Common Mistake to Avoid …

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We must place three vowels in three even positions and four consonants in four odd positions. The answer is $3! \times 4! = 6 \times 24 = 144$ .

The word ARTICLE has seven distinct letters: A, R, T, I, C, L, E. Three of these are vowels (A, I, E) and four are consonants (R, T, C, L).

When we arrange seven letters in a row, they occupy positions 1, 2, 3, 4, 5, 6, 7. The even positions are 2, 4, 6 (three slots), and the odd positions are 1, 3, 5, 7 (four slots). The constraint "vowels occupy the even place" means every vowel must sit in an even position, which forces every consonant into an odd position.

This is a two-stage placement problem. We have exactly three vowels for exactly three even slots, and exactly four consonants for exactly four odd slots — a perfect match on both sides.

Step-by-step reasoning:

Count the even and odd positions.

In a 7-letter arrangement, positions 2, 4, 6 are even (3 positions) and positions 1, 3, 5, 7 are odd (4 positions).
Identify vowels and consonants.

ARTICLE contains vowels {A, I, E} (3 letters) and consonants {R, T, C, L} (4 letters). All seven letters are distinct.
Place the vowels in even positions.

We need to arrange 3 distinct vowels in 3 distinct even slots. The number of ways to do this is the number of permutations of 3 objects, which is $3! = 6$ .
Place the consonants in odd positions.

Independently, we arrange 4 distinct consonants in 4 distinct odd slots. The number of ways is $4! = 24$ .
Apply the multiplication principle.

Since the placement of vowels and consonants are independent choices, the total number of arrangements is

$3! \times 4! = 6 \times 24 = 144.$ …

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Q.The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy the even place is (A) 144014401440 (B) 144144144 (C) 7!7!7! (D) 4C4×3C3{}^{4}C_{4} \times {}^{3}C_{3}4C4​×3C3​

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Q.The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy the even place is
(A) $1440$
(B) $144$
(C) $7!$
(D) ${}^{4}C_{4} \times {}^{3}C_{3}$

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