Question 1175160
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Problem 1


We require that 5 is in the units place. So that locks that digit up. The number looks like ABCD5, where A,B,C,D are the digit placeholders for some permutation of {2,3,4,6}. We aren't multiplying the A through D values out.


For slot A, we have four choices {2,3,4,6}
For slot B, we have three choices (one less than the number of choices in slot A)
For slot C, we have two choices (we count down one more)
For slot D, we have one choice


Overall, there's 4*3*2*1 = 24 different permutations here. Order matters.


If we don't fix that last digit to be a 5, then we have 5! = 5*4*3*2*1 = 120 different permutations.


The probability of getting a 5 in the units place is 24/120 = 1/5


Answer: <font color=red>1/5</font>


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Problem 2


Consider slots A,B,C,D,E where E is the units digit.


We want the five digit number to be even, so that must mean slot E takes on the value of 2, 4 or 6. We have three choices here.


Then for slot D, we have 5-1 = 4 choices after we ignore whatever is selected for slot E.
For slot C, we have 4-1 = 3 choices. We count our way down one at a time.
For slot B, we have 3-1 = 2 choices
For slot A, we have 2-1 = 1 choice


Multiply out those values: 3*4*3*2*1 = 3*4! = 3*24 = 72
There are 72 ways to form a five digit even number with the values {2,3,4,5,6} and repetition is not allowed. 


Divide this over 120, which was the number of ways to form a five digit number without restrictions, and we get 72/120 = 3/5


Answer: <font color=red>3/5</font>
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