If the letters of the word MASTER are permuted in all possible ways and the words thus formed are arranged in

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Q. If the letters of the word MASTER are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the rank of the word MASTER.

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Concept understanding — Permutations

Permutations: Arranging Things in Order

Imagine you have three different books — a Physics book, a Chemistry book, and a Maths book — and you want to place them on a shelf. How many different ways can you arrange them?

You could put Physics first, then Chemistry, then Maths. Or Physics, Maths, Chemistry. Or Chemistry, Physics, Maths. And so on. Each distinct ordering is called a permutation of the three books.

Note

A permutation is simply an arrangement of objects in a specific order. The key word is order — changing the order gives a different permutation.

Building the Intuition: The Multiplication Principle

Let's count the arrangements for those three books. For the first position on the shelf, you have 3 choices (any of the three books). Once you place one, for the second position you have 2 choices left. For the third position, only 1 choice remains.

So the total number of arrangements is:

3 \times 2 \times 1 = 6

This is the multiplication principle at work: the number of ways to do a sequence of tasks is the product of the number of choices at each step.

The General Formula

Now suppose you have $n$ distinct objects and you want to arrange all of them in a row. By the same logic:

1st position: $n$ choices
2nd position: $n-1$ choices
3rd position: $n-2$ choices
…
Last position: $1$ choice

The total number of permutations of $n$ distinct objects taken all at once is:

n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

This product has a special name — factorial — written as $n!$ (read "n factorial").

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

For example, $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ .

Important

By convention, $0! = 1$ . This is not a random rule — it makes formulas work consistently (there is exactly one way to arrange zero objects: do nothing).

What If You Don't Arrange All Objects?

Often you have $n$ objects but only want to arrange $r$ of them (where $r \leq n$ ). For instance, from 10 students, how many ways can you choose a first, second, and third prize winner?

1st prize: 10 choices
2nd prize: 9 choices
3rd prize: 8 choices

Total: $10 \times 9 \times 8 = 720$

In general, the number of permutations of $n$ distinct objects taken $r$ at a time is:

P(n, r) = n \times (n-1) \times (n-2) \times \cdots \times (n-r+1)

This product has exactly $r$ factors, starting from $n$ and decreasing. …

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Sort the letters alphabetically, and for each position count how many unused letters are alphabetically smaller than the actual letter there — each such letter would start a block of $(\text{remaining positions})!$ words that come before MASTER.

The word MASTER has 6 distinct letters. In alphabetical order:

$A,\ E,\ M,\ R,\ S,\ T$

Go position by position through M-A-S-T-E-R, at each step counting how many of the still-unused letters are alphabetically smaller than the letter actually placed there.

Position 1 — letter M. Unused letters: $\{A,E,M,R,S,T\}$ . Smaller than M: $A,E$ (2 letters). Each would fix a smaller letter first, giving $5!$ words before ours per choice:

$2\times5! = 2\times120 = 240$

Remove M. Remaining unused: $\{A,E,R,S,T\}$ .

Position 2 — letter A. Smaller than A among $\{A,E,R,S,T\}$ : none.

$0\times4! = 0$

Remove A. Remaining: $\{E,R,S,T\}$ .

Position 3 — letter S. Smaller than S among $\{E,R,S,T\}$ : $E,R$ (2 letters).

$2\times3! = 2\times6 = 12$

Remove S. Remaining: $\{E,R,T\}$ .

…

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Q.If the letters of the word MASTER are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the rank of the word MASTER.

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