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.
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%
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.
Please, if anyone can help and give a good explanation to back up their answer, I would greatly appreciate the help. Thank you!
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?
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Anyway I hope that you understand :)