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

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.

answer: your answer is a