-1
$\begingroup$

I need to find how many four digit numbers can be made using exactly 2 different digits. ALL DIGITS FROM 1 TO 9 CAN BE USED. NOT JUST 1 AND 0

  • 5
    You should make some effort yourself. Having people feed you answers here before you've even engaged your brain will not help you learn anything.2012-11-20

4 Answers 4

3

There are $\,2^4=16\,$ different ways to arrange two objects in patterns of $\,4\,$, from which you have to substract the patterns that contain all the elements equal, and there are $\,\binom{10}{2}=45\,$ different ways to choose two different objects out of $\,10\,$, so...

  • 0
    You meant "digits from 0 to 9"? Never mind: if it is only 1 to 9 then put 9 instead of 10 in my answer and change accordingly. The rest remains the same.2012-11-20
0

These are 2^4 it's same as having 4 bits represented by 0 and 1 if i understand you correctly.

0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 .... 1111

2^4 = 16

  • 2
    when 0000 and 1111 is not allowed then (2^4)-2 = 14 ?2012-11-20
0

Hint:

  • fixing a pair $(n,m)$ of digits, count the amuont of $4$-digit numbers you can make with them, using each of them at least one.
  • Count the number of of choices for $(n,m)$, and multiply by that. Should the order of the pair be taken into consideration?
  • 1
    @chndn: I think you mean all digits from 0 (not $1$) to $9$. And you did not need to post the comment here, as I was clearly aware of that. Indeed you only needed to post the comment to the answer by r4d1um (who apparently thinks in binary), which is the _only_ place you dit **not** do so. And you don't need to shout.2012-11-20
0

from the numbers $0$ to $9$, $2$ different numbers can be selected in ${10\choose2}$ ways. Of these, $9$ are $01$,$02$,$03$,$04$,$05$,$06$,$07$,$08$,$09$. ${10\choose2}=45.$ $45-9=36$ the groups are $12,13,14,15,16,17,18,19,23,24,25,26,27,28,29,34,35,36,37,38,39,45,46,47,48,49,56,57,58,59,67,68,69,78,79,89$ (different numbers).

Taking one group at a time, for example,$12$, we get $2^4$ numbers. therefore, for $36$ groups we get $36*(2^4)$ numbers. but, among these numbers there are numbers like $1111,2222,3333,4444,5555,6666,7777,8888,9999$ Each of these repeats $8$ times. So there are $8*9=72$ numbers with same digits.

Removing these, $36*(2^4)-72$ gives the number of numbers with two digits, not all digits same.

Now taking groups $01,02,03,04,05,06,07,08,09.$

Each group gives $2*2*2*1 =8$ numbers. In the multiplication I have put $1$ because digit in thousand's place cannot be $0$.

Among these numbers there are $1111,2222,3333,4444,5555,6666,7777,8888,9999$

So the final answer is $36*(2^4)-72+8.$