JEE Mathematics 2023 Question with Solution

Given: The digits are $1, 2, 3, 4, 5, 6, 7$.

Find: The number of permutations which contain neither the string $153$ nor the string $2467$.

Use the inclusion-exclusion principle.

Count permutations containing the string $153$ by treating $153$ as one block.

Then the objects are $153, 2, 4, 6, 7$, so the number of arrangements is

Count permutations containing the string $2467$ by treating $2467$ as one block.

Then the objects are $2467, 1, 3, 5$, so the number of arrangements is

Count permutations containing both strings $153$ and $2467$.

Then the two blocks are $153$ and $2467$, so the number of arrangements is

By inclusion-exclusion, the number of permutations containing at least one of these strings is

The total number of permutations of $7$ distinct digits is

Therefore, the number containing neither string is

So, the required number is $4898$.

The solution shows 154 in one place and 2367 in another, but the working values used are $5!$, $4!$, and $2!$, which correctly correspond to the question strings $153$ and $2467$. Hence the final answer remains $4898$.

Given: Total permutations of $1$ to $7$.

Find: Those with neither $153$ nor $2467$.

Treat each forbidden string as a single block.

Directly write:

Here, $5!$ counts arrangements containing $153$, $4!$ counts arrangements containing $2467$, and $2!$ is added back because arrangements containing both were subtracted twice.

Now compute:

Therefore, the required number is $4898$.