JEE Mathematics 2023 Question with Solution

Given: The letters of the word OUGHT are to be arranged in dictionary order.

Find: The serial number of the word TOUGH.

Arrange the letters in alphabetical order:

$$G, H, O, T, U$$

Now count all words that come before TOUGH in dictionary order.

Words beginning with letters before $T$ are:

$$G, H, O$$

For each such choice, the remaining $4$ letters can be arranged in

$$4!$$

ways. So the number of such words is

$$3 \times 4! = 72$$

Now consider words beginning with $T$.

For the second letter, letters before $O$ among the remaining letters $G, H, O, U$ are:

$$G, H$$

For each of these, the remaining $3$ letters can be arranged in

$$3!$$

ways. Thus the count is

$$2 \times 3! = 12$$

Now fix $TO$. The remaining letters are $G, H, U$. For the third letter, letters before $U$ are:

$$G, H$$

For each of these, the remaining $2$ letters can be arranged in

$$2!$$

ways. Thus the count is

$$2 \times 2! = 4$$

Now fix $TOU$. The remaining letters are $G, H$. For the fourth letter, letters before $G$ are none, so the count is $0$.

Therefore, the number of words before TOUGH is

$$72 + 12 + 4 = 88$$

Hence the serial number of TOUGH is

$$88 + 1 = 89$$

Therefore, the correct option is A.

The solution labels D as correct, but the extracted working evaluates to $89$, which matches option A.

Given: The word is TOUGH and the letters are arranged dictionary-wise.

Find: Its rank.

At each position, count how many smaller available letters could have been placed there, then multiply by the factorial of the remaining positions.

• Before $T$, there are $3$ smaller letters: $G, H, O$, contributing

$$3 \times 4!$$

• After fixing $T$, before $O$ there are $2$ smaller letters among the unused ones: $G, H$, contributing

$$2 \times 3!$$

• After fixing $TO$, before $U$ there are $2$ smaller letters among the unused ones: $G, H$, contributing

$$2 \times 2!$$

• After fixing $TOU$, before $G$ there are $0$ smaller letters, contributing $0$.

So the rank is

$$3 \times 4! + 2 \times 3! + 2 \times 2! + 1 = 89$$

Therefore, the correct option is A.