Question

(1) How many three-digit integers less than 701 have exactly two
different digits in their representation (for example, 232, or
466)?
a. 142 b. 146 C. 148
c. 148 d. 145
AS
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Answer 1

Given :

Three digit number less than 701.

To find :

Three digit number having exactly two different digits.

Solution :

In three digit number if exactly two digits are different then

In three digit number if exactly two digits are different thenexactly two digits are same.

So we have to find the numbers which has three digits and two digits are same.

To make it three digit number,

It can not have 0 as a digit a left most digit.

This number is less than 701, so

so

first digit having two same digits is 700.

So

From remaining numbers , it should be less than 700 now to find required numbers.

So,

Left most digit can have 1,2,3,4,5 and 6.

Middle and Right most digits can be 0 to 9 .

If digits has the type

AAB

then A at leftt will have any value from 1 to 6,

at middle will have same number so way get this number is 1.

and B at right can have any number other than choosen at left. so ways are = 9

so

Number of ways of digits as AAB :

$= 6 \times 1 \times 9 = 54$

similarly

Number of ways of digits as ABA :

$= 6 \times 9 \times 1 = 54$

If digits has type

BAA

then there are two choices, like :

If A = 1 to 6

then left most place B can not have that digit which is choosen as A.

So in this case ways of B = 5, ways of A at right= 6,

way of A at middle = 1 (digit selected as right)

so in this case number of ways :

$= 5 \times 1 \times 6 = 30$

If A is 0,7,8 and 9 (4 ways)

then left most place B can have all 6 values.

So in this case number of ways :

$= 6 \times 1 \times 4 = 24$

Total number of ways for given condition :

$= 54 + 54 + 30 + 24 + 1 = 163$

So total number of ways are 163.