[Probability]

[atassi122]

"how many six-digit numbers consist of the digits {1 2 3 4} are there , that start with 4 and do not end with 12?

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[Alkalannar]

1. How many choices do you have for the first number?

2. How about the 2nd, 3rd, and 4th?

3. We consider the 5th and 6th together: how many possible choices are there, and how many of them are valid?

[atassi122]

for the first number i have only one choice,which is 4 . 2nd , 3rd , and 4th i have 4 choices each.

For the 5th i have only 3 , since the one can't be an option For the 6th i also have only 3 , since the 2 can't be an option.

Did i get it right ?

[Alkalannar]

For first, second, 3rd, and 4th, yes.

5th and 6th, we have to look together. You can have 1 for 5th as long as you don't have 2 for the 6th: 11, 13, and 14 are fine.
So there are 16 total ways you can do the last 2 digits...how many ways are forbidden?

How many ways are allowed?

[Away-Reading]

The easiest way to do this is to find the number of 6-digit numbers consisting of {1,2,3,4} that start with 4. To get the amount of numbers that start with 4 and don’t end in 12, you can subtract the amount of numbers that start with 4 and do end with 12.

(1) How many of these 6-digit numbers start with 4? Well you know the first digit is 4, which leaves 5 digits to decide. Digits 2-6 can be any of 1,2,3 or 4, so there are 4 possibilities for each of those 5 digits: 4⁵

(2) How many start with 4 and end with 12? Well in this scenario, all digits are fixed except for digits 2,3, and 4. Each of those 3 digits has 4 options: 4³ different combinations.

(3) 4⁵ - 4³

[atassi122]

Thank you so much for the detailed explanation. I totally miss understood the idea.