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

Q: Why can't we just use $5^6$?

Using $5^6$ would count every possible 6-digit number, including those where some digits are missing (like 111111). The constraint 'all five digits must appear' is the key filter that $5^6$ ignores.

Q: Why do we divide by $2!$?

When you have identical items, swapping them doesn't create a new arrangement. Dividing by $2!$ removes these redundant counts, ensuring we only count unique numbers.

Q: Does the order of selection matter?

No, we are just selecting the 'repeating' digit. Whether we pick 3 first or 5 first doesn't change the set of digits we are arranging.

Question

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

Options:

(A) $5^6$

(B) $\frac{1}{2}(6!)$

(C) $6!$

(D) $\frac{5}{2}(6!)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To solve this problem, we must fill six slots using five distinct keys: $\{1, 3, 5, 7, 9\}$. Since we are required to use all five keys in six available slots, the **Pigeonhole Principle** dictates that exactly one digit must be repeated.

This creates a multiset of six digits where one digit appears twice and the other four digits appear once.

### Selecting the Repeating Digit

The first logical step is to determine which of the five available digits will be repeated. We must select exactly one digit from the set of five.

The number of ways to perform this selection is given by:

$$^5C_1 = 5$$

### The Arrangement Logic

Once the repeating digit is chosen, we have a set of six digits to arrange. For example, if we choose the digit $3$, our set becomes $\{1, 3, 3, 5, 7, 9\}$.

If all six digits were distinct, there would be $6!$ ways to arrange them. However, because we have two identical digits, we must divide by $2!$ to account for the indistinguishable permutations.

The number of distinct arrangements for any chosen multiset is:

$$\frac{6!}{2!} = \frac{720}{2} = 360$$

### The Final Synthesis

To find the total number of valid six-digit combinations, we multiply the number of ways to select the repeating digit by the number of ways to arrange the resulting multiset.

The calculation is as follows:

$$	ext{Total ways} = 5 	imes \frac{6!}{2!}$$

$$	ext{Total ways} = 5 	imes 360$$

$$\mathbf{	ext{Total ways} = 1800}$$

This result represents the complete set of valid arrangements, derived through the fundamental principles of combinatorics.

Animated Solution for Mathematics - Permutations and Combinations: Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is

Select Answer:

Visualized Solution

Analyze the Constraints

The Pigeonhole Principle

Selecting the Repeating Digit

Forming the Multiset

Arrangement Logic

Total Number of Ways

Final Simplification

The Sigma Insight: Linear Permutations

Analyzing the Setup

Selecting the Repeating Digit

The Arrangement Logic

The Final Synthesis

Similar Questions

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

The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is

Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are

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

The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 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 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

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