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

Q: Why do we split this into three cases? Can't we just use a single formula?

Great question! We split into cases because the set of remaining digits changes depending on which odd digit we fix in the unit place. If we fix 5, we have three 2s and two 3s left. If we fix 3, we have three 2s and one 3 left. Because the 'pool' of remaining digits changes, the denominator in our permutation formula changes, making separate cases necessary.

Q: Why do we divide by factorials like 3! and 2!?

That is the core of permutations with identical items! When you have repeating digits, swapping two identical digits (like two 2s) doesn't create a new, distinct number. Dividing by the factorial of the count of each repeated item 'divides out' these redundant arrangements, leaving you with only the unique, distinct numbers.

Q: What if the digit 0 was in the set? Would that change anything?

It absolutely would! If 0 were included, you would have to be careful about the first digit (the ten-lakhs place), as a 7-digit number cannot start with 0. You would then need to subtract the cases where 0 is in the first position, adding another layer of complexity to your combinatorial analysis.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a set of seven tiles: $\{1, 2, 2, 2, 3, 3, 5\}$. Your goal is to arrange these into a seven-digit odd number.

In the world of JEE Advanced, the secret to solving complex combinatorics is to identify the 'gatekeeper' constraint. Here, the constraint is simple: for a number to be odd, the unit digit must be odd.

### The Gatekeeper: Analyzing the Unit Digit

Our pool of odd digits is $\{1, 3, 5\}$. Because the composition of the remaining digits changes depending on which odd digit we place at the end, we must partition our problem into three distinct, mutually exclusive cases.

This is the art of systematic counting.

### Case 1: The Unit Digit is 5

If we lock the digit $5$ into the unit slot, we are left with the multiset $\{1, 2, 2, 2, 3, 3\}$. We have $6$ slots to fill.

We have $6$ items, but they are not all unique. The digit $2$ repeats $3$ times, and the digit $3$ repeats $2$ times. Using the formula for permutations of a multiset, the number of arrangements is given by:

$$
\frac{6!}{3! 2!} = \frac{720}{6 	imes 2} = 60
$$

### Case 2: The Unit Digit is 3

Now, we fix one of the $3$s in the unit slot. Our remaining pool is $\{1, 2, 2, 2, 3, 5\}$.

Notice the difference? We only have one $3$ left, while the digit $2$ still repeats $3$ times. The number of arrangements for this case is:

$$
\frac{6!}{3!} = \frac{720}{6} = 120
$$

### Case 3: The Unit Digit is 1

Finally, we fix the $1$ in the unit slot. Our remaining pool is $\{2, 2, 2, 3, 3, 5\}$.

This structure is identical to Case 1, with three $2$s and two $3$s. Thus, the calculation mirrors our first result:

$$
\frac{6!}{3! 2!} = 60
$$

### The Final Synthesis

We have explored every possibility. By the Rule of Sum, since these cases are mutually exclusive, we simply add them together to find the total number of valid seven-digit odd numbers:

$$
Total = 60 + 120 + 60 = 240
$$

There it is. Through careful partitioning and the application of multiset permutations, we have navigated the constraints to arrive at **240**.

Remember, in JEE, the math is rarely the hardest part—it is the clarity of your logic that wins the day.

Animated Solution for Mathematics - Permutations and Combinations: The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is ____.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Condition for Odd Numbers

Case 1: Unit Digit is $5$

Calculating Case 1

Case 2: Unit Digit is $3$

Calculating Case 2

Case 3: Unit Digit is $1$

Calculating Case 3

The Final Summation

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Gatekeeper

Case 1

Case 2

Case 3

The Final Synthesis

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