Question

This is a 3 part question. I have parts 1 and 2 correct. I need some help with the 3rd part please. :) The entry code for a house is five digits long, chosen from the digits 1-8. How many different codes are there?ANSWER: 32768 and THEN how many codes are there with no repeated digits? ANSWER: 6720
MY QUESTION: suppose that you have selected a code with no repeated digits at random. What is the probability that it alternates odd and even digits? (That is, you never have two odd or two even digits next to each other.) I answered this with .064 and was incorrect. My thinking was: I have a 50% chance of getting odd or even with the 1st selection, then I figured the EOEOE pattern was 288/6720 and the OEOEO was 144/6720 and I added those up to get 432/6720 ====Not Correct though

Answer 1

In the situation that no digit is repeated, the number of passcodes that alternate EOEOE is \(_4P_3\cdot\, _4P_2 = 4!\cdot \dfrac{4!}{2!} = 24*12 = 288\).

Likewise, the number of passcodes that alternate OEOEO is also \(_4P_3\cdot\, _4P_2 = 288\).

Hence, \(P(\text{passcode alternates between E & O}) = \dfrac{288}{6270}+\dfrac{288}{6270} = \dfrac{576}{6270}\).

Does this make sense? :-)

Answer 2

Yeah, I was thinking since the second pick had only 7 numbers to choose from 3 would be even and 4 would be odd....and the third pick had 6 numbers to choose from so only 3 could be even and 3 odd....I made it way more complicated and now that I look at it totally didn't make sense!

Thank you, I see it now. :)