Question

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

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{ How many 4 digit numbers are there whose decimal notation}\hspace{.33em}\\~\\
& \normalsize \text{ contains not more than 2 distinct digits ? }\hspace{.33em}\\~\\
& a.)\ 672 \hspace{.33em}\\~\\
& b.)\ 576 \hspace{.33em}\\~\\
& c.)\ 360 \hspace{.33em}\\~\\
& d.)\ 448 \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

B) 576 is the answer

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

sry to ask such a silly ques

Answer 7

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

Answer 8

http://prntscr.com/8kc5b1 :/

Answer 9

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 \(9\cdot \binom{3}{2}\)

Answer 10

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

Answer 11

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

Answer 12

wow why am I getting 765

Answer 13

@ganeshie8 !!!

Answer 14

Can you spot any mistake?

Answer 15

i think u double counted something

Answer 16

What?

Answer 17

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

Answer 18

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

Answer 19

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+

Answer 20

why would you change \(\binom{3}2\) to \(\binom{3}1\)

and good catch