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+3
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avatar+290 

If \(x%\) % of four-digit numbers have a repeated digit (the repeated digits do not need to be adjacent), then what is \(x\)? Express your answer as a decimal to the nearest tenth.

 Apr 26, 2018
 #1
avatar+99214 
+2

Well if one repeats and the others are both different and this is a code so the first digit can be zero THEN

 

Pick 3 numbers out of 10 = 10C3 = 120

these can be arranged in 3! =6 ways

 

120*6= 720 3 digit numbers.

There are 3 possibilities for the 4 digit 

and the 4th digit can go in 4 different places.

but there are 2 the same so must divide by 2

 

so we have

 

120*6*3*4/2 = 4320 ways I think

 

there are 10^4 = 10000 four digit numbers if the first one can be zeros

 

so the prob of chosing one with just one digit that repeats is  4320/10000 = 0.432 = 43.2%

x=43.2

 

If ithe number cannot start with 0 then it would be different.  

I can work this out but it is quite a bit more complicated.

 Apr 26, 2018
edited by Melody  Apr 26, 2018
 #2
avatar+290 
+5

I do not completely understand this... How would it be less?

TheMathCoder  Apr 26, 2018
 #3
avatar+290 
+1

Please, if anyone can help and give a good explanation to back up their answer, I would greatly appreciate the help. Thank you!

 Apr 27, 2018
 #4
avatar+99214 
+2

Did you understand any of it?

It would have been good if you had told me more specifically what you did not understand.

 

It is a 4 digit number but 1 digit (exactly one digit) is repeated just once.

So 3 digits out of 10 have to be chosen.

There are 10C3 = 120 ways to do this.

say the digits chose are  5,6 and 9, they can be in any order and there are 3! possible orders =6 orders.

So so far that is   120*6=720 ways to get a 3 digit number where the digits are all different.

 

But we want a 4 digit number.

One digit has to be repeated there are 3 possibilities so that is   720*3 = 2160 posibilities.

 

But there are 4 postions that extra number can be put in. So that is  2160*4=8640 numbers

But I have double counted because    1434 is the same if I swap the 4s around so I have to halve this answer

 

8640/2 = 4320    

So the number of favourable outcomes is 4320.

 

Altogether there are  10*10*10*10 = 10000 4 digit numbers (if they are allowed to start with 0)

 

so the prob of getting just one digit repeated once is   4320/10000 = 0.4320 = 43.2%

 

Do you understand now?

You will need to write your query on this thread but also send me a private message with the address of this thread attached. Otherwise I will not see it. 

 

If you do have a query try to ask a specific question.  

Anyway I hope that you understand :)

 Apr 27, 2018
 #5
avatar+99214 
0

Hi Mathcoder,

 

I see that you have viewed my answer, I am glad about that.    smiley

But you have not given me any written feedback.                        sad

 

I did ask specific questions of you. It would be nice of you to respond. 

 Apr 28, 2018

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