Counting

Question

+1

494

2

Of all the four-digit positive integers containing only digits from the set {2,4,6,8} what fraction of them have at least one 8?

Guest May 22, 2021

WillBillDillPickle · Answer 1 · 2021-05-22T02:44:22

3 * 3 = 9

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We choose which diget is 4.

Guest · Answer 2 · 2021-05-23T01:49:05

The total number of 4 digit numbers with digits from the set given is $(4)(4)(4)(4) ={4}^{4}.$ If we leave out 8, the total number changes to

$(3)(3)(3)(3)={3}^{4}$ . So the fraction having at least one 8, which is 1 minus the fraction having no 8, would then be

$1-\frac{{3}^{4}}{{4}^{4}}=1-\frac{81}{256}=\frac{175}{256}$