How many 4 digit numbers are there whose decimal notation contains not more than 2 distinct digits ?

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How many 4 digit numbers are there whose decimal notation contains not more than 2 distinct digits ?

Mathematics
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\(\large \color{black}{\begin{align} & \normalsize \text{ How many 4 digit numbers are there whose decimal notation}\hspace{.33em}\\~\\ & \normalsize \text{ contains not more than 2 distinct digits ? }\hspace{.33em}\\~\\ & a.)\ 672 \hspace{.33em}\\~\\ & b.)\ 576 \hspace{.33em}\\~\\ & c.)\ 360 \hspace{.33em}\\~\\ & d.)\ 448 \hspace{.33em}\\~\\ \end{align}}\)
B) 576 is the answer
how do we know you are right? a straight answer like this is no use to anyone please explain how you derived it, or better still give the poster the method , and let HIM derive it

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Note to self: No silly mistakes. No silly mistakes. No silly mistakes.
well what do u mean by decimal notation here :/ sry u posted one more question like decimal notation i got stuck there too
sry to ask such a silly ques
decimal notaion ={0, 1,2,3,4,5,6,7,8,9,} i think
http://prntscr.com/8kc5b1 :/
If there is only one unique digit then there are 9 such numbers (1111, 2222, ...) If there are two unique digits: First, consider all numbers where zero does not appear. (i) Three of one digit and one of the other one. \(9\cdot \binom{4}{3} \cdot 8\) (ii) Two of each digit \(9\cdot \binom{4}{2}\cdot 8\) Now consider those numbers where zero does appear. (i) Three zeroes and one other digit. This can only be done in 9 ways (1000, 2000, ...) (ii) Two zeroes and two of the other digit. This can be done in \(9\cdot \binom{3}{2}\)
I think "decimal notation" means nothing but base 10 integers here.
(iii) One zero and three of other digits.\[3\cdot 9~ways \]
wow why am I getting 765
Can you spot any mistake?
i think u double counted something
What?
I'm actually getting 765 + 27 = 792 :P
Slightly modified : If there is only one unique digit then there are 9 such numbers (1111, 2222, ...) If there are two unique digits: First, consider all numbers where zero does not appear. (i) Three of one digit and one of the other one. \(9\cdot \binom{4}{3} \cdot 8\) (ii) Two of each digit \(9\cdot \dfrac{\binom{4}{2}}{\color{red}{2}}\cdot 8\) Now consider those numbers where zero does appear. (i) Three zeroes and one other digit. This can only be done in 9 ways (1000, 2000, ...) (ii) Two zeroes and two of the other digit. This can be done in \(9\cdot \binom{3}{2}\) (iii) One zero and three of the other digit. This can be done in \(\color{red}{9\cdot \binom{3}{1}}\) @ParthKohli
adding them all gives 576 http://www.wolframalpha.com/input/?i=9%2B9*%5Cbinom%7B4%7D%7B3%7D*8+%2B+9*%5Cdfrac%7B%5Cbinom%7B4%7D%7B2%7D%7D%7B2%7D*8%2B9%2B9*%5Cbinom%7B3%7D%7B2%7D+%2B+9*%5Cbinom%7B3%7D%7B1%7D+
why would you change \(\binom{3}2\) to \(\binom{3}1\) and good catch

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