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Detailed solutions of math problems: Problem #1

Problem: How many of the natural numbers from 1 to 1000 have none of their digits repeated?

Solution:

In order to solve this problem, you have to consider different digit numbers separately. There are one digit, two digit and three digit natural numbers from 1 to 1000. We will calculate the number of one digit, two digit and three digit numbers from 1 to 1000 that have none of their digits repeated separately. Keep in mind that there are only ten digits that can make up a number, those from 0 to 9.

One digit numbers:
There are nine one-digit natural numbers, viz. 1, 2, 3, 4, 5, 6, 7, 8 and 9. Since there is only one single digit in these numbers, therefore these nine numbers have none of their digits repeated.

Two digit numbers:
Consider a two digit number represented by two empty boxes as follows:
There can be nine different digits (1 to 9) put in the first box above, and ten different digits (0 to 9) put in the second box above in order to form a two digit number. Zero can not be put in the first box, otherwise it will become a single digit number. Thus we have a total number of 9 choices for the first box and ten choices for the second box. Therefore total number of different two digit numbers that can be formed are 9 x 10 = 90. You can verify this by manually counting all two digit numbers from 10 to 99. You can see the tree chart (sample space) below to visualize it:

But, the 90 different two digit numbers obtained above include numbers like 66, in which the digits are repeating. So we will reconsider our above logic and correct it as follows:

There can be nine different digits (1 to 9) put in the first box above. In the second box, you can put 10 different digits (0 to 9), but if you do not want repeating digits, you will not choose a digit for the second box that is there in the first box. Therefore, for each digit chosen for the first box, only nine different digits can be chosen for the second box excluding the digit already chosen for the first box. Thus, the total number of two-digit numbers that have no repeating digits are 9 x 9 = 81.

Three digit numbers:
Consider a three digit number represented by the box below:
  • The first box can be filled with digits 1 to 9. (9 choices)
  • Second box can be filled with digits 0 to 9 excluding the digit taken in the first box. (9 choices)
  • Third box can be filled with digits 0 to 9 except the two digits taken above (8 choices) 
Total number of permutations = 9 * 9 * 8 = 648.
Hence there can be 648 different three digit numbers which do not have any repeating digit.

Only single digit, two digit, and three digit numbers will be considered, since you have to find the required numbers till only 1000, which is the first four digit number.

Total number of numbers from 1 to 1000 that have none of their digits repeated = 9 + 81 + 648 = 738.

Therefore there are 738 numbers from 1 to 1000 that have none of their digits repeated.

Solved!

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