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How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and all digits are different from each other?

Given:
The number is a 3 digit number
The first digit is different from zero
All digits are different from each other

Each digit of the 3 digit number can be any number from 0 - 9 (total 10 digits)

Digit in Hundred's place can we any digit from 1-9 = 9 options
Digit in ten's place can have any digit from 0 - 9 , except the digit in 100's place = 9 options
Digit in unit's place can have any digit from 0-9 except the 2 digits already in 10s and 100s = 8 options

Total possible numbers = 9*9*8 = 648

Ans: C
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There is one condition in the first slot so we fill that one first.

Other than 0, there are 9 choices.

9 x 9 x 8 = 648

Answer: C
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let the places be HTU
for H -> 9 ways
for T -> 9 remaining
and for U -> 8 remaining
so total = 9*9*8

that is
648

C is the answer
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