All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The

Q: Wait, why do we select positions for the odd digits first? Can't we just arrange them directly?

That is a fantastic question! If you try to arrange them directly, you might get confused because you have $4$ even positions but only $3$ odd digits. By selecting the positions first using $^4C_3$, you are essentially creating a 'blueprint' for the number. Once you have chosen the $3$ specific slots for the odd digits, the remaining $6$ slots are automatically reserved for the even digits. It simplifies the problem into two independent, manageable tasks.

Q: Why do we divide by factorials like $2!$ or $4!$?

Great catch! We divide by factorials whenever we have identical items. For example, in the set of odd digits $\{1, 1, 3\}$, the two $1$s are indistinguishable. If you swapped the two $1$s, the number would look exactly the same. Dividing by $2!$ removes these redundant, identical arrangements, ensuring we only count unique numbers.

Q: Is the order of multiplication important? Does it matter if I arrange the even digits first?

Mathematically, no! Multiplication is commutative, so $4 \times 3 \times 15$ is the same as $15 \times 4 \times 3$. However, logically, it is often easier to handle the most constrained part of the problem first—in this case, the odd digits—to build your mental model of the number's structure.

Question

All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is :

Options:

(A) $175$

(B) $162$

(C) $160$

(D) $180$

Sigma Percentile · Accepted Answer

### The Architecture of Numbers: A Combinatorial Journey

Welcome, fellow traveler, to the elegant world of combinatorics. Today, we are not just solving a problem; we are building a structure.

Imagine you have nine empty pedestals in a row, waiting to be filled by a specific set of digits: $\{1, 1, 2, 2, 2, 2, 3, 4, 4\}$. Our mission is to arrange these digits such that the odd ones—the $1$s and the $3$—are strictly confined to the even-numbered pedestals.

Let us break this down into a logical, step-by-step masterclass.

### Phase 1: The Inventory

Before we place a single digit, we must know our materials. We have a collection of nine digits, which we categorize by their nature:

- **Odd Digits**: $\{1, 1, 3\}$ (Total: $3$)
- **Even Digits**: $\{2, 2, 2, 2, 4, 4\}$ (Total: $6$)

This categorization is our foundation. Our nine pedestals are numbered $1$ through $9$. The even-numbered pedestals are $2, 4, 6,$ and $8$, meaning there are exactly $4$ such pedestals available.

### Phase 2: The Selection Strategy

Here is where the intuition kicks in. We have $4$ even pedestals, but only $3$ odd digits.

We must choose which $3$ of these $4$ pedestals will host our odd digits. This is a classic selection problem using the combination formula $^nC_r$:

$$^4C_3 = \frac{4!}{3!(4-3)!} = 4$$

We have $4$ distinct ways to choose the pedestals for our odd digits. This is our first building block.

### Phase 3: The Arrangement of the Odd

Now that we have selected our $3$ pedestals, we must place the odd digits $\{1, 1, 3\}$ into them.

Because we have two identical $1$s, we must account for this repetition by dividing by the factorial of the repetition count:

$$	ext{Arrangements} = \frac{3!}{2!} = \frac{6}{2} = 3$$

So, for every selection of pedestals, there are $3$ unique ways to arrange the odd digits.

### Phase 4: The Even Digits and the Grand Finale

With the odd digits settled, we look at the remaining $6$ pedestals. We have $6$ even digits left: $\{2, 2, 2, 2, 4, 4\}$.

The digit $2$ appears $4$ times, and the digit $4$ appears $2$ times. The number of ways to arrange these is given by the multinomial coefficient:

$$\frac{6!}{4!2!} = \frac{720}{24 	imes 2} = \frac{720}{48} = 15$$

We have $15$ ways to arrange the even digits. Now, we bring it all together. Since these steps are independent, we multiply the number of ways for each step:

$$	ext{Total Ways} = (	ext{Selection of pedestals}) 	imes (	ext{Arrangement of odds}) 	imes (	ext{Arrangement of evens})$$

$$	ext{Total Ways} = 4 	imes 3 	imes 15 = 180$$

And there you have it! By carefully selecting our positions and then arranging our digits while respecting their identities, we have found that there are exactly **$180$** such numbers.

Remember, in combinatorics, always secure your constraints first, and the rest will fall into place with elegant precision.

Animated Solution for Mathematics - Permutations and Combinations: All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is :

Select Answer:

Visualized Solution

Analyze the Given Digits

Categorize Odd and Even Digits

Visualize the Nine Positions

Identify the Even Positions

Select Positions for Odd Digits

Calculate the Selection

Arrange the Odd Digits

Calculate Odd Arrangements

Remaining Digits and Places

Arrange the Even Digits

Calculate Even Arrangements

Calculate Total Numbers

Final Answer

The Sigma Insight: Linear Permutations

The Architecture of Numbers

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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

The number of three-digit even numbers, formed by the digits 0, 1, 3, 4, 6, 7 if the repetition of digits is not allowed, 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 ____.

The sum of all the 4-digit distinct numbers that can be formed with the digits 1,2,2 and 3 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 ?

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

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

The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to ____.