Question

25) How many 3 digits numbers are formed from digits (0 to 9) have exactly one digit "2" in the
number in any place?
(Repetition is allowed)
(3 Points)
A) 225
B) 240
0 120
D) 160​

Answer 1

Answer:

Method 1:

The number is three digits, so for them let's take three blanks _ _ _

The first blank can be filled using any of the digits from 1–9 because if we use zero to fill the first blank the number becomes of two digits. Hence we have 9 ways to fill the first blank.

Now, the second blank can be filled by any of the remaining 10 digits because repetition is allowed and thus the digit selected for the first blank can also be selected. So 10 ways.

Similarly 10 ways for the third blank.

So total number of combinations become 9 x 10 x 10 = 900

Hence the answer is 900 such number can be formed.

Method 2:

Since the first digit cannot be zero, we have 9C1 ways to select the first digit (one digit selected from a set of nine distinct digits). (9C1 = 9)

Now, for the remaining two places we can have zero as well. Hence we have 10C1 ways to select a digit for tens and ones place each. (10C1 = 10)

Hence total number of combinations become 9C1 x 10C1 x 10C1 = 9 x 10 x 10 = 900

Hence the answer is 900 such numbers can be formed.

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