How many natural numbers not exceeding 4321 can be formed with the digits 1,2,3 and 4,if the digits can repeat?

The answer is 229.I just need the steps.

Dear Student

Here is the answer to your question:The given digits are 1, 2, 3 and 4. These digits can be repeated while forming the numbers. So, number of required four digit natural numbers can be found as

Consider four digit natural numbers whose digit at thousandths place is 1.

Here, hundredths place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Similarly, tens place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Ones place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Number of four digit natural numbers whose digit at thousandths place is 1 = 4 × 4 × 4 = 64

Similarly, number of four digit natural numbers whose digit at thousandths place is 2 = 4 × 4 × 4 = 64

Number of four digit natural numbers whose digit at thousandths place is 3 = 4 × 4 × 4 = 64

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**Now, consider four digit natural numbers whose digit at thousandths place is 4:**

Here, if the digit at hundredths place is 1, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 2, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 1, then ones place can be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 2, then ones place can be filled only in 1 way so that the number formed is **not** exceeding 4321.

Number of four digit natural numbers **not** exceeding 4321 and digit at thousandths place is 3 = 4 × 4 + 4 × 4 + 4 + 1 = 37

Thus, required number of four digit natural numbers **not** exceeding 4321 is 64 + 64 + 64 + 37 = 229.

Cheers!

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