The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain

Q: Wait, why don't we multiply by $3!$ for the '153' block?

That is a fantastic question! We would multiply by $3!$ if the problem asked for the digits 1, 5, and 3 to be *together* in any order. However, the problem specifically says the *string* '153' must appear. This means the internal order is fixed as 1-5-3. Therefore, there is only $1$ way to arrange the digits inside that block.

Q: Why do we subtract the intersection $n(A \cap B)$?

Think of it as a Venn diagram. When we add the number of permutations containing '153' ($n(A)$) and the number of permutations containing '2467' ($n(B)$), we have counted the cases where *both* strings appear twice—once inside $n(A)$ and once inside $n(B)$. To get the correct count of the union, we must subtract that overlap once.

Q: Is there a way to solve this without the Principle of Inclusion-Exclusion?

While you could try to manually list cases, it is extremely error-prone and inefficient for JEE Advanced. The Principle of Inclusion-Exclusion is a powerful, systematic tool designed exactly for these types of 'at least one' or 'neither/nor' problems. It is the most elegant and reliable path to the solution.

Question

The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is _______ .

Sigma Percentile · Accepted Answer

### Analyzing the Total Universe

First, let us look at the big picture. We have seven distinct digits: $\{1, 2, 3, 4, 5, 6, 7\}$.

If there were no restrictions, the number of ways to arrange these digits is given by the permutation of $n$ distinct objects, which is $n!$.

For our seven digits, the total number of unrestricted permutations is $7!$:

$$7! = 7 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1 = 5040$$

This value represents our total universe of possibilities.

### The 'Tie' Method for Constraints

Now, let us tackle the restrictions. The problem forbids the string '153' and the string '2467'.

To handle a constraint where a group of digits must stay together, we use the **'Tie Method'**. Imagine taking the digits 1, 5, and 3 and tying them together into a single, unbreakable block, $B_1$.

Now, instead of seven individual digits, we have $B_1$ and the remaining digits $\{2, 4, 6, 7\}$. This gives us five items to arrange:

$$n(A) = 5! = 120$$

Since the internal order of '153' is fixed, we do not need to multiply by any internal permutations.

We apply the same logic to the second constraint. We tie 2, 4, 6, and 7 into a single block, $B_2$.

We are left with $B_2$ and the digits $\{1, 5, 3\}$, totaling four items to arrange:

$$n(B) = 4! = 24$$

### The Elegant Dance of PIE

We have calculated the 'bad' cases for each constraint, but we must account for the overlap. If a permutation contains *both* '153' and '2467', we have counted those cases in both $n(A)$ and $n(B)$.

This is where the **Principle of Inclusion-Exclusion (PIE)** comes to our rescue. We need to find the number of permutations containing *both* strings, denoted as $n(A \cap B)$.

In this scenario, we treat '153' as block $B_1$ and '2467' as block $B_2$. Since all seven digits are accounted for in these two blocks, we are simply arranging two items:

$$n(A \cap B) = 2! = 2$$

Now, we use the PIE formula to find the total number of 'bad' permutations:

$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$

$$n(A \cup B) = 120 + 24 - 2 = 142$$

These 142 permutations are the ones that contain at least one of the forbidden strings.

### The Final Victory

We are almost there! We know the total number of permutations is 5040, and the number of 'bad' permutations is 142.

To find the number of valid permutations, we subtract the bad cases from the total:

$$	ext{Required} = 5040 - 142 = 4898$$

By breaking the problem down into manageable blocks and applying the Principle of Inclusion-Exclusion, we have navigated through the complexity to find the exact answer of **4898**.

Animated Solution for Mathematics - Permutations and Combinations: The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is _______ .

Enter Numerical Value:

Visualized Solution

Understanding the Digits

Total Permutations

The Strategy: PIE

Counting Set $A$ : The '153' Block

Calculating $n (A)$

Counting Set $B$ : The '2467' Block

Calculating $n (B)$

The Overlap: $n (A \cap B)$

Calculating $n (A \cap B)$

Applying PIE Formula

Final Subtraction

The Sigma Insight: Linear Permutations

Analyzing the Total Universe

The 'Tie' Method for Constraints

The Elegant Dance of PIE

The Final Victory

Similar Questions

The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to__________

The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is ____.

Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is

The number of 9 digit numbers, that can be formed using all the digits of the number 123412341 so that the even digits occupy only even places, is ____.

Total number of six-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is

The number of integers, greater than 7000 that can be formed, using the digits 3, 5, 6, 7, 8 without repetition, is

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is

How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions ?

A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done is

The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is