How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. Match each item in Column $C_1$ with its correct answer in Column $C_2$. $C_1$: (a) $4$ letters are used at a time; (b) All letters are used at a time; (c) All letters are used but the first is a vowel. $C_2$: (i) $720$; (ii) $240$; (iii) $360$.

> [!TLDR] > This problem is about counting permutations of distinct letters from the word MONDAY. The key idea: treat each case as arranging a subset of 6 distinct items, with or without a restriction on the first letter. The matches are: (a) → 360, > (b) → 720, > (c) → 240. The word MONDAY has 6 distinct letters: M, O, N, D, A, Y. No letter repeats, so every arrangement is a permutation without repetition. The core concept is simple: when you choose $r$ distinct objects from $n$ and arrange them in order, the number of ways is $P(n, r) = \frac{n!}{(n-r)!}$. For all 6 letters, it's just $6!$. Let’s work through each part. 1. **Part (a): 4 letters used at a time** We are selecting any 4 of the 6 distinct letters and arranging them in a sequence. The number of such 4-letter words is the number of permutations of 6 items taken 4 at a time: $$P(6,4) = \frac{6!}{(6-4)!} = \frac{720}{2} = 360.$$ So (a) matches with (iii) 360. 2. **Part (b): All letters used at a time** This is simply arranging all 6 distinct letters in every possible order. That’s $6! = 720$. So (b) matches with (i) 720. 3. **Part (c): All letters used but the first is a vowel** The vowels in MONDAY are O and A — two vowels. The restriction: the first letter must be a vowel. - Step 1: Choose the first letter. There are 2 choices (O or A). - Step 2: Arrange the remaining 5 letters in any order in the remaining 5 positions. That’s $5! = 120$ ways. Total = $2 \times 120 = 240$. …

How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming

NCERT Exemplar · Q64

C_1

(a)

4

letters are used at a time;

(b) All letters are used at a time;

C_2

(i)

720

;

(ii)

240

;

(iii)

360

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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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This problem is about counting permutations of distinct letters from the word MONDAY. The key idea: treat each case as arranging a subset of 6 distinct items, with or without a restriction on the first letter. The matches are: (a) → 360,

(b) → 720,

The word MONDAY has 6 distinct letters: M, O, N, D, A, Y. No letter repeats, so every arrangement is a permutation without repetition. The core concept is simple: when you choose $r$ distinct objects from $n$ and arrange them in order, the number of ways is $P(n, r) = \frac{n!}{(n-r)!}$ . For all 6 letters, it's just $6!$ .

Let’s work through each part.

Part (a): 4 letters used at a time We are selecting any 4 of the 6 distinct letters and arranging them in a sequence. The number of such 4-letter words is the number of permutations of 6 items taken 4 at a time:

P(6,4) = \frac{6!}{(6-4)!} = \frac{720}{2} = 360.

So (a) matches with (iii) 360.

Part (b): All letters used at a time

This is simply arranging all 6 distinct letters in every possible order. That’s $6! = 720$ .

So (b) matches with (i) 720.
Part (c): All letters used but the first is a vowel

The vowels in MONDAY are O and A — two vowels. The restriction: the first letter must be a vowel.
- Step 1: Choose the first letter. There are 2 choices (O or A).
- Step 2: Arrange the remaining 5 letters in any order in the remaining 5 positions. That’s $5! = 120$ ways. Total = $2 \times 120 = 240$ . …

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Q.How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. Match each item in Column C1C_1C1​ with its correct answer in Column C2C_2C2​. C1C_1C1​:

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Q.How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. Match each item in Column $C_1$ with its correct answer in Column $C_2$ . $C_1$ :

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