Question 393532: Suppose 5 different integers are randomly chosen from between 20 and 69, inclusive. What is the probability that they each have a different tens digit?
Found 2 solutions by edjones, richard1234:
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! There are 5 different sets of tens digits, 2,3,4,5,6 and there are 10 of each of them.
I will assume that they are chosen without replacement.
nCr=combination of n things taken r at a time.
(10C1*10C1*10C1*10C1*10C1)/50C5
=100000/2118760
=.0472
.
Ed
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! There are 50 integers in the set, so 50C5 possible ways to choose 5 different integers. Also, there are 5 possible tens-digits, 2 through 6, so each of the five numbers must represent one of these tens digits. Since there are 10 ways to choose a number with a given tens digits (as in 30, 31, ... 39) there are 10^5 total ways to choose 5 numbers that satisfy these properties.
The probability would therefore be (10^5)/(50C5).
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