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. 40 pts here!!

Answer

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

e=61

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