We need the dictionary rank of the word MONDAY among all permutations of its letters.

The letters are all distinct: $A, D, M, N, O, Y$ There are $6! = 720$ total words.

We arrange them in alphabetical order and count how many words come before MONDAY.

Step 1: First letter = $M$

Alphabetical order of letters: $A < D < M < N < O < Y$

Before words starting with $M$, we have words starting with $A$ and $D$.

For each fixed first letter, remaining $5$ letters can be arranged in: $5! = 120$ ways.

So words before those starting with $M$: $2 \times 120 = 240$

Step 2: Second letter = $O$

Now fix first letter as $M$. Remaining letters are: $A, D, N, O, Y$

We want second letter $O$. Letters smaller than $O$ among remaining are: $A, D, N$ So there are $3$ such letters.

For each choice, remaining $4$ letters can be arranged in: $4! = 24$ ways.

Hence number of words before $MO\cdots$ among words starting with $M$: $3 \times 24 = 72$

Running total: $240 + 72 = 312$

Step 3: Third letter = $N$

Now fix $MO$. Remaining letters are: $A, D, N, Y$

We want third letter $N$. Letters smaller than $N$ are: $A, D$ So there are $2$ such letters.

For each, remaining $3$ letters can be arranged in: $3! = 6$ ways.

Count: $2 \times 6 = 12$

Running total: $312 + 12 = 324$

Step 4: Fourth letter = $D$

Now fix $MON$. Remaining letters are: $A, D, Y$

We want fourth letter $D$. Letters smaller than $D$ among remaining: $A$ So there is $1$ such letter.

For each, remaining $2$ letters can be arranged in: $2! = 2$ ways.

Count: $1 \times 2 = 2$

Running total: $324 + 2 = 326$

Step 5: Fifth letter = $A$

Now fix $MOND$. Remaining letters are: $A, Y$

We want fifth letter $A$. There is no letter smaller than $A$.

Count added: $0$

Running total: $326$

Step 6: Sixth letter = $Y$

Only one letter remains, so no extra count.

Thus, number of words before MONDAY is: $326$ So its serial number is: $326 + 1 = 327$

Option check

• A: $324$ ❌
• B: $328$ ❌
• C: $326$ ❌
• D: $327$ ✅

Therefore, the serial number of MONDAY is 327.