How many three-digit integers less than ^@ 601 ^@ have exactly t | Math Question and Answer

Answer:

$116$

Step by Step Explanation:

Let the two different digits be $x$ and $y.$
Therefore, the required integers are of the form $xxy, xyx$ or $yxx.$
If the repeated digits are zero, we must ignore the form $xxy, xyx$ as they will give us one and two digit numbers. $Eg. 001, 010,$ etc.
So, if $x = 0,$ the integers have the form $yxx$ and $y$ can be $1, 2, 3, \ldots , 6.$
Therefore, there are $6$ integers with two zeros, $i.e. 100, 200, \ldots, 6 00.$
When the repeated digit is non-zero, the integers are of the form $xxy, xyx$ or $yxx.$
If $x = 1, y$ can be $0, 2, 3, 4, 5, 6, 7, 8$ or $9,$ therefore there are $9 \times 3$ $= 27$ possible integers but we must ignore $011$ as this is a two-digit integer.
Since your number is less than $601$ so we must ignore $611, 711, 811, 911.$
This gives $27 - 5 = 22$ different integers.
Similarly, there will be an additional $22$ integers for every non-zero value of $x$ .
Therefore, the total number of three-digit integers less than $601$ that have exactly two different digits in their representation $= 6 + (5 \times 22) = 116.$
Hence, there are $116$ three-digit integers less than $601$ that have exactly two different digits in their representation.

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