Question 1175160: A five-digit number is formed by arranging 2, 3, 4, 5, and 6 randomly without repetition. Find the probabilities of the following events
1. The unit digit of the number is 5.
2. T he number is an even even number.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
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: 1/5
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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: 3/5
Answer by ikleyn(52855)  (Show Source):
You can put this solution on YOUR website! .
The solution by @math_tutor is perfectly correct.
I want to show you another way of thinking.
Having 5 different digits, you can form 5! = 1*2*3*4*5 = 120 different numbers.
In this soup of numbers, the last digit can be any of the given 5 digits with the same probability (obviously !).
Therefore, the amount of such numbers with any fixed last digit is  of the total amount of the numbers.
It means that the probability under the problem question is . ANSWER
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Thinking this way, you may solve other similar problems VERY QUICKLY without making unnecessary calculations.
But you should be aware about APPLICABILITY of this way thinking (!) - it depends on the problem (!)
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