Number of Natural numbers UPTO ONE LAKHS which contain 1,2,3 exactly once and remaining digit anytime

To determine how many natural numbers up to one lakh (100,000) contain the digits 1, 2, and 3 exactly once, while allowing the remaining digits to be any valid digit, we need to consider a few key points about the structure of natural numbers and the constraints given.

Understanding the Problem

Firstly, we need to clarify what it means for a number to contain the digits 1, 2, and 3 exactly once. This means that in our natural number, the digits 1, 2, and 3 must appear one time each, and the remaining digits (if any) can be any digit from 0 to 9, but they cannot include 1, 2, or 3. Additionally, we are looking for numbers that have a maximum of five digits since we are limited to numbers up to 100,000.

Case Analysis Based on the Number of Digits

We can break this down based on the number of digits in the natural numbers:

• 3-digit numbers
• 4-digit numbers
• 5-digit numbers

Counting 3-Digit Numbers

For a 3-digit number, the only digits are 1, 2, and 3. The number of ways to arrange these three digits is simply the permutation of three distinct items:

3! = 6

So, we have 6 valid 3-digit combinations: 123, 132, 213, 231, 312, and 321.

Counting 4-Digit Numbers

For a 4-digit number, we must include 1, 2, and 3 exactly once, plus one additional digit. The additional digit cannot be 1, 2, or 3, so it can be any digit from 0-9 except for these three. Thus, we have the digits {0, 4, 5, 6, 7, 8, 9} available to choose from, giving us 7 options.

Now, we can select one digit from these 7 options, and there are:

7 choices for the additional digit.

After selecting this digit, we can arrange the four digits (1, 2, 3, and the chosen digit) in:

4! = 24 ways.

Thus, the total number of 4-digit numbers is:

7 * 24 = 168

Counting 5-Digit Numbers

For a 5-digit number, we need to include 1, 2, and 3 exactly once, plus two additional digits. These additional digits can be chosen from the same set of {0, 4, 5, 6, 7, 8, 9}, which again consists of 7 digits. We must consider that we can choose the same digit more than once.

We can choose any two digits from the available set. The number of combinations of selecting two digits (with repetition allowed) can be computed using the formula for combinations with repetition:

The number of ways to choose 2 digits from a set of 7 is given by:

C(n + r - 1, r) = C(7 + 2 - 1, 2) = C(8, 2) = 28.

After selecting these two digits, we can arrange 5 digits (1, 2, 3, and the two chosen digits) in:

5! = 120 ways.

Thus, the total number of 5-digit numbers is:

28 * 120 = 3360

Summing Up the Counts

Now that we have calculated the valid numbers for each case, let's add them together:

For 3-digit numbers: 6

For 4-digit numbers: 168

For 5-digit numbers: 3360

Therefore, the total number of natural numbers up to one lakh that contain the digits 1, 2, and 3 exactly once, with the remaining digits being any valid digit, is:

6 + 168 + 3360 = 3534

So, the final answer is that there are 3534 natural numbers up to one lakh containing the digits 1, 2, and 3 exactly once.