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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The interquartile range of the players' weights = 48 pounds.
The boxplot attached to this question is missing. It was obtained online and is attached to this solution of the question.
It should be noted that the following is true for a boxplot.
A box plot gives a visual representation of the distribution of the data, showing where most values lie and those values that greatly differ from the rest, called outliers.
The elements of the box plot are described thus;
The bottom side of the box represents the first quartile, and the top side, the third quartile. Therefore, the width of the central box represents the inter-quartile range.
The horizontal line inside the box is the median.
The lines extending from the box reach out to the minimum and the maximum values in the data set, as long as these values are not outliers. The ends of the whiskers are marked by two shorter horizontal lines.
Variables in the dataset, higher than Q3+(1.5×IQR) or lower than Q1-(1.5×IQR) are considered outliers and are usually shown using dots above the top whisker or below the bottom whisker.
So, it is evident that for this question,
First quartile = 174 pounds
Third quartile = 222 pounds
Inter Quartile Range = (Third quartile) - (First quartile)
= 222 - 174
= 48 pounds.
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