How many integers from 1 through 100,000 contain the digit 6 exactly once?
1 answer:
Consider the set of all (not-all-zero) decimal strings of length 6. This is the set of strings
000001
000002
...
099998
099999
100000
There are obviously 100,000 strings in this set, so we have a one-to-one correspondence to the integers between 1 and 100,000. Think of any string starting with 0s as the number with the leading 0s chopped off.
There are two choices for the first digit, either 0 or 1, but a number can only contain a 6 if the first digit is 0; otherwise, the number would exceed 100,000. For every digits place afterward, if a given digits place contains a 6, then the remaining four places have 9 possible choices each, choosing from 0-9 excluding 6. If we fix the 6 in, say, the second digits place, then the number of integers between 1 and 100,000 containing exactly one 6 is
where the first 1 refers to the only choice of 0 in the first digits place, the second 1 refers to the unique 6 in the next place, and the remaining four places are filled with one of 9 possible choices.
Now, notice that we can permute the digits of such a number in 5 possible ways. That is, there are 5 choices for the placement of the 6 in the number, so we multiply this count by 5.
000001
000002
...
099998
099999
100000
There are obviously 100,000 strings in this set, so we have a one-to-one correspondence to the integers between 1 and 100,000. Think of any string starting with 0s as the number with the leading 0s chopped off.
There are two choices for the first digit, either 0 or 1, but a number can only contain a 6 if the first digit is 0; otherwise, the number would exceed 100,000. For every digits place afterward, if a given digits place contains a 6, then the remaining four places have 9 possible choices each, choosing from 0-9 excluding 6. If we fix the 6 in, say, the second digits place, then the number of integers between 1 and 100,000 containing exactly one 6 is
where the first 1 refers to the only choice of 0 in the first digits place, the second 1 refers to the unique 6 in the next place, and the remaining four places are filled with one of 9 possible choices.
Now, notice that we can permute the digits of such a number in 5 possible ways. That is, there are 5 choices for the placement of the 6 in the number, so we multiply this count by 5.
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Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).
<h3>What is a confidence interval of proportions?</h3>A confidence interval of proportions is given by:
In which:
is the sample proportion.
- z is the critical value.
- n is the sample size.
In this problem, we have a 95% confidence level, hence, z is the value of Z that has a p-value of
, so the critical value is z = 1.96.
We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:
Hence the bounds of the interval are found as follows:
The 95% confidence interval is (0.2316, 0.3112).
More can be learned about the z-distribution at brainly.com/question/25890103