If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning

Q: Why do we add $1$ at the end of the calculation?

The rank is the position, not the count of words *before* it. If you count all words that come before 'MOTHER', you have the count of preceding words. Adding $1$ accounts for the word 'MOTHER' itself, which is the very next word in the sequence.

Q: What if the word had repeating letters, like 'APPLE'?

Great question! If letters repeat, the number of permutations changes. Instead of $n!$, you would use the formula $\frac{n!}{p!q!...}$ where $p, q$ are the frequencies of repeating letters. The logic remains the same, but the counting becomes more nuanced because you must divide by the factorial of the frequency of each repeating letter.

Q: Why do we fix letters one by one instead of just calculating the whole thing at once?

Combinatorics is about breaking down complexity. By fixing letters, we reduce the problem from a $6$-letter permutation to a $5$-letter, then $4$-letter, and so on. It turns a massive, intimidating problem into a series of small, manageable steps that mirror how a dictionary is actually structured.

Question

If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning) be listed as in a dictionary, then the position of the word 'MOTHER' is

Sigma Percentile · Accepted Answer

### The Dictionary Journey: Decoding 'MOTHER'

Welcome, future engineer! Today, we are going to decode the secret language of dictionaries. Have you ever wondered how a dictionary knows exactly where to place a word? It is not magic; it is the beautiful, rigid logic of combinatorics.

We are going to find the rank of the word 'MOTHER'. Think of this as finding a specific coordinate in the vast, ordered space of all possible permutations of these six letters.

### Phase 1: The Alphabetical Foundation

Before we can start counting, we must understand the rules of the game. The dictionary lists words in strict alphabetical order.

Our first step is to take the letters of 'MOTHER'—$M, O, T, H, E, R$—and arrange them in their alphabetical hierarchy: $E, H, M, O, R, T$. This is our map.

Every word we encounter in our search will be compared against this order. We have six empty slots to fill, and we are going to systematically count every word that comes before 'MOTHER'.

### Phase 2: The Systematic Elimination

We start by looking at the first slot. If we fix $E$ in the first position, we have $5$ remaining letters to arrange in $5$ slots.

$$5! = 120 	ext{ words}$$

Since 'MOTHER' starts with $M$, all $120$ words starting with $E$ come before ours. We add $120$ to our tally.

Next is $H$. Again, fixing $H$ in the first slot gives us $5! = 120$ words. These also come before 'MOTHER'.

Our tally is now $120 + 120 = 240$. We are making progress!

Now, we reach $M$. This matches the first letter of 'MOTHER', so we lock $M$ in the first slot. We move to the second slot. The alphabetical order of the remaining letters is $E, H, O, R, T$.

We need $O$ for the second slot, but we must first count all words starting with $ME$ and $MH$. For $ME$, we have $4$ remaining slots, giving us $4! = 24$ words. For $MH$, we have another $4! = 24$ words.

Our tally is now $240 + 24 + 24 = 288$.

### Phase 3: The Final Stretch

We have reached $MO$. This matches the second letter of 'MOTHER', so we lock $O$ in the second slot. Now we look at the third slot.

We need $T$, but alphabetically, we must pass $E, H,$ and $R$ first. For each of these three letters, we have $3$ remaining slots, giving us:

$$3 	imes 3! = 3 	imes 6 = 18 	ext{ words}$$

Our tally is now $288 + 18 = 306$.

Finally, we reach $MOT$. This matches the third letter of 'MOTHER', so we lock $T$ in the third slot. Now for the fourth slot, we need $H$.

Alphabetically, the first available letter is $E$. Words starting with $MOTE$ have $2$ remaining slots, giving us $2! = 2$ words. Our tally is now $306 + 2 = 308$.

We are almost there! The next letter for the fourth slot is $H$, which matches! We lock $H$. We have $E$ and $R$ left.

Alphabetically, $E$ comes before $R$, so the very next word is 'MOTHER'. We add $1$ for the word itself.

The final rank is **$308 + 1 = 309$**.

Animated Solution for Mathematics - Permutations and Combinations: If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning) be listed as in a dictionary, then the position of the word 'MOTHER' is

Enter Numerical Value:

Visualized Solution

Introduction to Dictionary Rank

Words Starting with $E$

Words Starting with $H$

Words Starting with $M$ and $M E$

Words Starting with $M H$

Words Starting with $M O$ and $M O E, M O H, M O R$

Words Starting with $M O T$ and $M O T E$

Reaching the Word $M O T H E R$

Final Rank Calculation

The Sigma Insight: Linear Permutations

The Dictionary Journey

Phase 1

Phase 2

Phase 3

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