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mathmath333
 one year ago
How many 4 digit numbers are there whose decimal notation
contains not more than 2 distinct digits ?
mathmath333
 one year ago
How many 4 digit numbers are there whose decimal notation contains not more than 2 distinct digits ?

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{ How many 4 digit numbers are there whose decimal notation}\hspace{.33em}\\~\\ & \normalsize \text{ contains not more than 2 distinct digits ? }\hspace{.33em}\\~\\ & a.)\ 672 \hspace{.33em}\\~\\ & b.)\ 576 \hspace{.33em}\\~\\ & c.)\ 360 \hspace{.33em}\\~\\ & d.)\ 448 \hspace{.33em}\\~\\ \end{align}}\)

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0how 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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Note to self: No silly mistakes. No silly mistakes. No silly mistakes.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0well what do u mean by decimal notation here :/ sry u posted one more question like decimal notation i got stuck there too

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0sry to ask such a silly ques

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0decimal notaion ={0, 1,2,3,4,5,6,7,8,9,} i think

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2If 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 \(9\cdot \binom{3}{2}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I think "decimal notation" means nothing but base 10 integers here.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2(iii) One zero and three of other digits.\[3\cdot 9~ways \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2wow why am I getting 765

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Can you spot any mistake?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i think u double counted something

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I'm actually getting 765 + 27 = 792 :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Slightly 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3adding 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+

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2why would you change \(\binom{3}2\) to \(\binom{3}1\) and good catch
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