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

1. mathmath333

\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}}

2. Drigobri

3. MrNood

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

4. ParthKohli

Note to self: No silly mistakes. No silly mistakes. No silly mistakes.

5. imqwerty

well what do u mean by decimal notation here :/ sry u posted one more question like decimal notation i got stuck there too

6. imqwerty

sry to ask such a silly ques

7. mathmath333

decimal notaion ={0, 1,2,3,4,5,6,7,8,9,} i think

8. imqwerty
9. ParthKohli

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 $$9\cdot \binom{3}{2}$$

10. ParthKohli

I think "decimal notation" means nothing but base 10 integers here.

11. ParthKohli

(iii) One zero and three of other digits.$3\cdot 9~ways$

12. ParthKohli

wow why am I getting 765

13. ParthKohli

@ganeshie8 !!!

14. ParthKohli

Can you spot any mistake?

15. mathmath333

i think u double counted something

16. ParthKohli

What?

17. ParthKohli

I'm actually getting 765 + 27 = 792 :P

18. ganeshie8

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

19. ganeshie8
20. ParthKohli

why would you change $$\binom{3}2$$ to $$\binom{3}1$$ and good catch